Solar Cookout!

Hi everyone,

Here is the form that you can fill out to say what you can bring tomorrow:



TO view what others have posted, please go to THIS link:

For suggestions on materials, click on the links below. I'm pretty sure we can get a solar oven together! (We may have some plastic in the robotics room, as well as the copy paper boxes and insulating paper at school. The rest can probably be found if we look carefully at home!

http://www.instructables.com/id/The-5-6-Solar-Oven/
http://www.instructables.com/id/Plastic_Bottle_solar_oven/

Page magnifiers are sold at Staples, Office Max, Barnes and Noble. Having more than one would make the cooking part go MUCH more quickly....

Remember to email, call, or text when you have an idea of what you want to bring, and I can post it to the form myself, if you aren't near a computer.

I can see clearly now!

Solution to O5 exam:


Almost there guys. Hang tough!

When Life Gives You lemons, make a battery....

I want to thank you all for the work you are putting in - it is making a difference, and I am proud of how far you have all come since the start of our review!

Solutions for yesterday's 2003, and the ninety 8 for the weekend.

Here is the worksheet you can use for predicting your score on the 1998 exam.

Give it a try and let me know what you think - I have tried to work out ALL bugs, but I may have missed one....

It's Not Easy being Green...(When Earth Day is Long Gone)

2006 solutions are here.

Just a little bit closer now...

Kinetic Theory/Thermo Solutions:
1 B
2 B
3 A
4 E
5 C
6 C
7 C
8 D
9 E
10 A
11 E
12 A
13 C
14 D
15 A
16 A

'03 solutions, and 1993 solutions

I Know These Constants Are Correct, by Wally Sheep

(Anybody know the book by Wally Lamb? I guess it's a little obscure....)

Here is a link to the table of constants given on the exam. Use it as you do the part II problems.

Abbot's Surprise: A Magnetic Personality!

1. a
2. b
3. c
4. a
5. a
6. c
7. b
8. e
9. b
10. b
11. a
12. a
13. c
14. a
15. e.
16. b
17. d
18. a
19. c
20. e

And, for your viewing and char-broiled rubric-flavored enjoyment, the rubric for the 1994 exam.

What do you Quantum?

1D
2A
3D
4C
5C
6B
7B
8B
9C
10C
11D
12A
13A
14D
15E
16A
17E
18A

1981B1
a) F = Ffric = 20 N
b) Wnet = Fnet*delta_x*(1) = 60 J, so the net force is 15 N. Since the friction force is 20 N, this means the new force F' = 20 N + 15 N = 35 N.
c) The net force acting on the block is 15 N, and the block has a mass of 10 kg. The acceleration is then a = Fnet /m = 1.5 m/s^2.

2.
a) The work done to compress the spring is equal to the final kinetic energy of the block.

KE = 1/2mv^2 = Wspring = 150 J
b) When the blocks are held together, the energy stored in the spring is again 150 J. Thus the kinetic energy when the blocks are moving apart is 150 J.

1/2*M1*v1^2 + 1/2*M2*v2^2 = 150 J

Conservation of momentum also applies:

total initial P = total final P

0 = m1*v1 - m2*v2

The system can be solved for v1 = 15 m/s, v2 = 5 m/s

3.
a) Fe to the right, gravity down, T diagonally to the left at 30 degrees from the vertical.
b) T cos(30) - mg = 0 so T = .058 N
T sin(30) - Felec = 0 and Felec = qE so E = 5.8 x 10³ N/C
c) Since there is a net force in both the downwards direction (due to gravity) and the horizontal direction (due to the electric force), the ball will follow a straight path down and to the right.


4.
total Req = 8 ohm, total emf in the circuit is (60 V - 12 V = 48 V, notice that the second battery is backwards!)

This means that the current in the circuit is 48 V / 8 ohms = 6 Amperes.

a) V across the parallel combination is 6 V, so the current through the 2 ohm resistor is 3 A.

b) P = I^2*R = (6 A)^2 * (3 ohms) = 108 W

c) Because of the direction of current, the battery is being charged. The terminal voltage is the battery voltage plus the voltage added by the resistor going backwards.

V = 12 V + 6A*1 ohm = 18 V

5.
a) Ray diagram:

b) The image is real

c) Use the thin-lens equation equation and solve for di = 9 cm

d)


On this part, you can use the trick suggested in the image, OR you can use the image as the object for lens 2. The important thing is to NOT change how the rays are drawn relative to the primary and secondary foci. The first ray enters the lens parallel and towards the left, and is "brought towards" the primary focus.

The only weird thing is that since the primary focus is on the same side as the "object", the light ray is bent as if it comes from the primary focus. I can show this in class tomorrow.

Oh Say, can you c...(where c is the speed of light)

1 D
2 D
3 E
4 E
5 E
6 E
7 B (n1 = 1.2/.8)
8 C
9 D
10 E
11 D (convex mirror is similar to a diverging lens, in that it always produces a virtual image of an object)
12 D (this is a magnifying glass! remember that converging lenses can form both real and virtual images)
13 D
14 D (be sure to draw BOTH the reflected and refracted ray. The angle of the reflected ray will always be 40 degrees from the normal, which means the angle from the surface is 50 degrees. Depending on the index of refraction of the material, the refracted ray will be at most 40 degrees from the vertical, which is an angle of 50 degrees from the surface. Thus the total angle will be between (40 + 50) to (40 + 90 degrees)
15 C

1995B2.
a) R = V^2/P = 24 ohms
b) Energy used = P*delta_t = 6W * (30 days*24 hrs/day*3600 seconds/hr) = 1.6 x 10⁷ J
c) The bulb should be in series with the resistor, and the two should be in parallel with the battery terminals. The toaster should be connected in parallel with the battery so it gets 120 V across it.
d) The bulb needs 12 V across it, which means the resistor must have 108 V across it to add to 120 V. The power of the bulb is 6 W, which by P = IV, means the current must be 0.5 A. Therefore the resistance must be V/I = 108 V/0.5 A = 216 ohms
e)
i. If the variable resistor is increased, the current through the branch will decrease. Since the resistance of the bulb remains the same, the power P = I^2*R will decrease, and the bulb will get dimmer.

ii. The toaster has the same amount of voltage across it, regardless of the resistor. the power will therefore stay the same.

1995B3
a) T directed up, mg down, so T - mg = 0 in equilibrium, and T = mg = 1 N

b)
T sin(th) = Th = ma = (0.1 kg)(5 m/s^2) = 0.5 N
T cos(theta) - mg = 0 with Tv = T cos(theta) = mg = 1 N

c) Constant speed! in equilibrium.

T - mg = 0, so T = Tv = mg = 1 N.
Th = 0 since there are no components of T in the x direction.

d)
Th = ma_x = (0.1 kg)(5 m/s^2)*sin(30) = 2.5 N
Tv = ma_x...

I will post more later, I just realized I am late for an appointment!

Wave/Interference Review

Waves/Interference Review Solutions:
1 D
2 A
3 C
4 C
5 B
6 A
7 D
8 E
9 C
10 E
11 C
12 D
13 A
14. B (the radical symbol should end before the f)
15. B
16. A
17. B
18. A
19. D

Preview for tomorrow...let's shed some light on MC optics questions....

Working Hard, or hardly working?

Hi everyone,

I want to thank those of you that have contacted me thus far. Those of you that have not...it's cranky time....(Azeezat, Maria, Danielle, Brandon, Sam, Marlon...did you lose my number?)

I've been working hard in Cleveland to get wedding details figured out - running around the city, picking things up, dropping them off, and of course, deciding on cake...here's the evidence of my "hard work" after dinner tonight:




Actually, I'm feeling a bit sick. I never thought I could have too much cake.

Thankfully, you can NEVER have enough physics.

Take care everyone,

EMW

So You Think You Can Physics?

Hi everyone,

I hope you are enjoying your time away from school...I just wanted to give you guys a link to the rubrics for the 1986, 1987, and 2007 exams.

Click on the date and save the file to your computer.

Be SURE to contact me and let me know how things are going!

Is this really the end? I was just getting into it....

Problem 1.
a + b)


Don't worry about the fact that the signs are different from what we discussed in class today - your answers to the other parts are still consistent with the solutions below, and will be correct.
c) Highest energy is the 6 eV transition.
d) lowest is the 1 eV transition.
e) This is the reverse of c and d - longest wavelength will be the 1 eV transition since E = hc/lambda
f) shortest wavelength is with the 6 eV transition, which has the highest energy.
g) The difference in energy between these two energy levels is 5 eV. This translates into a photon of wavelength 248 nanometers.
h) Three are possible.
i) The electron has KE = 7.12 x 10^-19 J = 4.45 eV. This is enough energy to cause an n = 1 to n = 3 transition, but not n = 1 to n = 4.

Problem 2.

1996B5.
a) Z = 102

b) K = 1/2*mv^2 = 8.42 MeV = 1.35 x 10^-12 J
alpha particle mass = (4 u)*(1.66 x 10^-27kg
v = (2*K/m)^1/2 = 2.02 x 10⁷ m/s

c) A correct explanation includes a reference to:
1. a change in mass/mass defect being converted into KE fo the alpha particle

2. conservation of energy - binding energy of the nucleus is converted into KE of the alpha particle during the decay.

3. work-energy theorem - the electric force does work on the alpha particle as it moves away from the nucleus, thereby changing its KE.

Nuke it - Mr. Weinberg has no Gas Stove

Chapter 30:

Q1. Isotopes of a single element have the same number of protons, so they act the same way in a chemical reaction as each other. They have different numbers of neutrons, so the mass of isotopes vary.

Q2 & 3.
a) Uranium-232, 92 protons, 140 neutrons
b) Nitrogen-13, 7 protons, 6 neutrons
c) Hydrogen, 1 proton, 0 neutrons
d) Strontium-82, 38 protons, 44 neutrons
e) Berkelium-247, 97 protons, 150 neutrons

Q10.
Gamma rays are photons or electromagnetic waves, so they have no mass and no charge. Alpha particles have a relatively large mass, a positive charge, and high energy. beta particles have a relatively small mass, a negative charge, and also high energy.

Weinberg Problems:
1.
a) 2.42 m/s
b) 0.642 Hz
c) 3.91 N

2.
a) 225 ohms
b) 75 ohms

3.
a) By the right hand rule, positive charges moving west in the Earth's magnetic field (directed North) will experience a downwards force, so the bottom part of the tube will become positive.
b) 200 m/s (pretty fast!)


4.
a) Green light has a wavelength between 500 and 550 nanometers, so any value in this range is a good estimate.
b) Using c = f*lambda and lambda = 525 nm, f = 5.7 x 10¹⁴ Hz
c) Frequency is the same, wavelength becomes 525 nm/ 1.4 = 375 nm.
d) The path difference is 2t, and the wave inverts twice upon reflection because the index of refraction of the reflecting surfaces is larger in both cases.

Thus 2t = m*lambda, and for the lowest nonzero thickness, t = 187.5 nm.

Are you here from the Wave side, or the Particle side of the family?

1.
a) 2.8 eV
b) 1.7 x 10¹⁸ photons
c) 5.5 x 10⁵ m/s
d) 1.3 x 10^-9 m


2.
a) 6.38 x 10^-10m
b)335 nm
c) The diagram consists of three transitions, 1.2, 3.7, and 4.9 eV.



3.
a) The best fit curve is shaped like a cosine curve.
b) 1.408 x 10^-14 m
c) 4.71 x 10^-20 kg*m/s
d) 8.08 x 10^-14 J

4.
a) 1.33 x 10¹⁵ Hz
b) 276 nm (not visible), 497 nm, 621 nm
c) Because of the energy of the photons in the given range, transitions can occur from the ground state to either the 1st or 2nd excited states.

5.
a)122 nm
b)The energy is the difference between the ionization energy and the energy of the second photon.
13.6eV - 10.2 eV = 3.4 eV
c)8.2 x 10¹⁴Hz
d)
1995B5.

I Emitted the Photon...YOU just didn't absorb it...

1997B Part II.
a)
b) The energy must be -1.9 eV, based on the energy emitted by the 600 nm photon.
c) 1240 nm - outside of the visible spectrum


200...Something, #7:
a) There are actually two (correct) ways to draw the diagram:

b) 1240 nm, again, outside of the visible spectrum.


7.
a) 9.9 x 10^-15 J
b) 3.3 x 10^-23 kg*m/s
c) The photon wavelength increases because some of the energy of the incident photon was lost to KE of the electron.
d) 6.0 x 10^-23 kg*m/s

1993B6
a)1.69 x 10¹⁹ Hz
b)3.73 x 10^-23 kg*m/s
c)3.31 x 10^-16 J
d)2.46 x 10^-23 kg*m/s

That Smudge on the X-ray is from my Jelly Donut.

Photoelectric Effect
http://www.lon-capa.org/~mmp/kap28/PhotoEffect/photo.htm

More visual Photoelectric effect:
http://phet.colorado.edu/simulations/photoelectric/photoelectric.jnlp

Solutions:
2000B5
a)
i) 4.5 eV or 7.2 x 10^-19 J
ii) 1.26 x 10⁶ m/s
b) 183 nm
c) 5.56 x 10¹⁴ Hz

1980B3
a) The graph should be straight forward, but you SHOULD know that all four points cannot be on the same line!
b) 0.75 x 10¹⁴ Hz
c) 3.1 eV
d) Energy of electrons is 5 eV according to data, so the stopping potential will be 5 volts.
e) Charges move in circular paths when they are in a magnetic field perpendicular to their velocity. This is a possible reason for this behavior.

Page 855:
16. 4.14 x 10¹⁴ Hz

17. 401 nm

19. The shortest visible wavelength photons have energies of 3.1 eV (400 nm wavelength). Copper and Iron have work functions that are greater than this, so they will not eject photo electrons in visible light.
20.

a) 3.49 x 10^-19 J or 2.179 eV.
b) 0.930 V

Page 885:
P32. 4.14 x 10^-11 m (Don't worry about the longest wavelength.)
P33. 41,000 V

From Handout:
1.
a) 2.5 mm
b) wavelength = 385 nm, frequency = 6 x 10¹⁴ Hz
c) Since the wavelength decreases, and the spacing between maxima is proportional to wavelength, the space between the maxima will decrease.

1991B6.
a) 3.9 x 10^-5 m
b) 9.6 x 10^-3 m

Problems:
7. 8.5 x 10^-5m
10. The important thing here is to notice that the ratio d/L is the same for both. You should obtain the equation that 2*lambda_1/y = m*lambda_2/2*y, with y the same for both since the question asks for the wavelength at which a minimum appears at the same location. Solving for a wavelength lambda_2 in the visible spectrum, lambda_2 = 613 nm
20. Solve for theta using sin(theta)= m*lambda/W, and then use the fact that L*tan(theta) = height above the center. The distance is 1.591 meters.
21. Notice that this says the distance between maxima, not minima. The path difference must be half a wavelength more than that required for the minimum. The total angle is 19.5 degrees across the central maximum, so the angle between the central maximum and the first minimum is 9.75 degrees. sin(9.75) = 1.5*lambda/W, so W = 5606 nm.

There's this thin film on my teeth...(Eww)

Hi everyone,

Here is the link to the physics lesson website - be sure to check out the lesson on thin film interference, as well as on diffraction.

Please also be aware that there will be construction Saturday on the 6 line downtown on the way to the REACH workshop at Baruch College. Please give yourself plenty of time to get there by 8:00 AM.

From handout:

4. thickness = 1.05 x 10^-7 m

5. thickness = 96.1 nanometers

1984B5
a) 5 x 10¹⁴ Hz
b) 4.8 x 10^-7 m
c) 1.2 x 10^-7 m
d) 2.4 x 10^-7 m

2000B5.

I apologize for not getting this solutions up quickly, but you must be aware that part (a) of this question is worth 8 of the 15 points. You need to make sure you draw BOTH angles of reflection, BOTH angles of refraction (after using Snell's Law), and include the angle of incidence (35 degrees) of the light ray upon hitting the bottom surface of the glass.

Parts b and c are, by comparison, relatively tame. I will post this ASAP.


38. wavelength = 643 nm, which is colored red.
40. 169 nm
43. 27*lambda/n = 2t, so t = 9045 nanometers, or 9.045 x 10^-6 m
45. 2t = n*(640 nm/1.36), and 2t = (n + 1/2)*512 nm/1.36. Solving the system, t = 471 nanometers.

It's Critical that you get the right angle!

Solutions to the AP Problems:
79B6.
a) Remember that you MUST use the angles relative to the normal vector - n1 sin(t1) = n2 sin (t2)
n2 = 1.327
n = c/v so v = 2.26 x 10⁸ m/s

b) If you try to use Snell's law at point Q, you will find you end up trying to take the inverse sine of a number greater than 1 - not possible. This means the ray will NOT refract, but instead reflect. This is total internal reflection. You can justify the fact that it reflects either by calculating the critical angle and showing that the 53 degrees is greater than the critical angle (49 degrees), or by showing that when refraction does not occur, reflection does.

c) If the lens was made of plastic in air, it would be a converging lens. Since it is the reverse, it will be a diverging lens, and the rays will spread out.

88B5.
a) The ray will be refracted slightly downwards from the horizontal dotted line.

b)
The ray will pass directly through the left interface because the incident angle is 0. At the right interface, the angle from the normal is 37 degrees. By Snell's law, the refracted ray will have an angle of 64.5 degrees.

Be careful though, as the question asks for the angle from the horizontal. To get this, you must subtract the 37 degrees from the 64.5 degrees, giving a final answer of 27.5 degrees.

c) Here you must assume that the incident angle of 37 degrees is the critical angle.

By Snell's law, theta_1 = 37, n2 = 1.0 (for air), and theta_2 = 90 since we are assuming total internal reflection occurs. Solving, n1 = 1.667

d) Here the light ray will again be refracted below the horizontal line, but the angle should not be as great as it was for part (a).

e) Using Snell's law again with n1 = 1.667, theta_1 = 37 degrees, and n2 = 1.33, we can solve for theta_2 = 49.0 degrees. You must again subtract 37 degrees from this value to obtain the angle from the horizontal of 12 degrees below.

93B4.
a) v = c/n = 1.875 x 10⁸ m/s
b) There are two ways to do this - the fast way is to remember the equation from Monday that the wavelength in the material lambda' = lambda/n where lambda is the wavelength in a vacuum. 700 nm/1.5 = 467 nm.

We can also use c = f*lambda to find the frequency of the red light in a vacuum:
f = 3E8 m/s/(700 E-9 m) = 4.28 x 10¹⁴ Hz

The frequency stays the same when a wave changes from one material to another, so this is also the frequency in the glass. We also know the speed of red light in the glass from part a.

lambda = v/f = 1.875 x 10⁸ m/s/(4.28 x 10¹⁴ Hz) = 438 nm (either answer would be acceptable)

c) Frequency can be found the same as before - make SURE before you use the v = f*lambda equation, you are using the speed and wavelength for the same medium!

f = 3E8 m/s/(700 E-9 m) = 4.28 x 10¹⁴ Hz

d) The key here is to notice that the incident rays have the SAME initial angle hitting the right side of the prism (30 degrees), as well as the same n2 = 1 (for air.) The difference is in n1.

From Snell's Law:
n1 sin(theta_1) = n2*sin(theta_2)

sin(theta_2) = n1*sin(theta_1)

Since n1 is greater for the blue light, theta_2 will be greater for the blue light. This means that the blue light will bend further away relative to the normal than the red.

e) In this case, it is a similar situation when the rays reach the right surface (the rays again do not refract at the left surface because the incident angle is zero.) The issue here, however, is that n1 is the same for both (n1 = 1), and theta_1 is again 30 degrees.

Snell's law again:

n1 sin(theta_1) = n2*sin(theta_2)

sin(theta_2) = 1*sin(theta_1)/n1

In this case, since n1 is greater for blue light, it will refract less
than with the red light. Thus, the blue light will bend above the red light, with both rays bending towards the normal.

#3.
b) Using the thin lens equation, 1/di = 1/f - 1/do, so di = 30 cm. This should roughly match your answer for part (a).

c) hi/ho = -di/do so hi = 2*5cm = 10 cm

d) This can be a tricky one. If you use what we came up with today, you will find that the image from the first lens is on the right side of the second lens. You can use the rules as we always do, except that the primary focus of the second lens is now on the same side as the object, so the lens will act like a diverging lens.

The slick way to get this is to notice that Ray 2 leaves the first lens parallel to the axis when it enters the second lens. This ray enters lens 2 parallel, so it is brought to the primary focus of lens 2. Ray 1, after passing through lens 1, passes through the center of lens 2, which means it will pass through undeflected. This will result in these two rays converging at a single point to the right of lens 2. This is the location of the image, roughly 7 cm to the right of lens 2.

e) The image is inverted and smaller than the original object.

Can you Lens me a dollar? Times are tough....

Multiple Choice problems from the handout:
1. B

2. E

3. C

4. D


Questions:
20. If the object distance is very large, then by the thin lens equation, 1/d0 approaches zero. This means the image distance is equal to the focal length. This is where the film must be placed to form an image.

Problems:
48b. 390 mm

50. Since the image is on the other side of the lens, it must be a real image. This means that the image distance is positive. Using the thin lens equation, the focal length must be 39.8 cm.

52. 72 cm on the same side of the lens as the stamp (virtual image, negative image distance). Magnification = -4

60.
a) 15.2 cm, real image (opposite side of the lens as object), height is -2.78 mm
b) -12.1 cm, virtual image (negative image distance), height is +2.22 mm

83. Since a diverging lens has a negative focal length, by the thin lens equation, 1/di can never be positive. If di is negative, the image is on the same side as the object, so it will be virtual. The only way for di to be positive is if d0 for the object is negative. This can occur, and we will discuss HOW it can occur in class tomorrow!

AP Problems:
1989B6.
a) real
b) di = 30 cm
c) M = -0.5
d) See the image to the left.
e) focal length decreases - by Snell's law, the rays will be bent more as they enter and exit the lens, so the place where the rays intersect the principal axis will be closer to the center of the lens.

1997B5.

a) converging - the image is located behind the lens, which means image distance is positive. This can only occur with a converging lens.

b) 1/f = 1/di + 1/do so f = 22.5 mm

c) The rays will probably intersect near the 90 mm mark, but the 22.5 mm focal length makes it difficult to get it exactly. Don't worry if it isn't exactly on the 90 mm mark.

d) The image is inverted, real, and larger than the object. You can also confirm the size change by noticing that M = -di/do = -90/30 = -3.

e) The resulting image should be inverted and to the left of the mirror.

Come on Ray, just give me the equation!

A reminder: Please try the simulations below. They will go a long way to giving you an intuitive understanding of how the light rays move in both the cases of refraction and reflection. If you want a preview of tomorrow's lesson, try the lenses in the bottom simulation.

Index of Refraction simulator: http://www.ps.missouri.edu/rickspage/refract/refraction.html

Site for Simulating Mirrors (lenses too, coming soon to a 6/7 period near you!)
http://www.surendranath.org/Applets/Optics/ReflRefrCurv/CurvSurfApplet.html


Question 15 - The fish is, in reality, lower in the aquarium than is seen by the eye in the picture. Remember that the light ray from the fish is bent away from the normal when it leaves the water. This means that the actual ray from the fish to the water's surface is tilted downwards to reduce the angle from the normal to the surface.

Problems:
12. image is virtual, upright, and reduced in size. It is located at 2.09 cm behind the surface of the ball. (d_i = -2.09 cm).

13. magnification is -di/d0, so you can find di using the magnification of 3 (positive because it is upright.) Solving for the focal length, and multiplying by 2, the radius R = 3.9 meters.

14. concave mirror, R = 5.66 cm

27. n = c/v = 1.310

34. The light emerges from the first interface at an angle of 27.7 degrees. By geometry, we can figure out the angle at which this hits the right side of the prism. A quadrilateral is formed by the top vertex of the prism, the point where the refracted ray hits the left side, the right side intersection point, and the intersection of the normal vectors. (See the picture below)


We can see that this ray hits the other surface of the prism at 33 degrees. By Snell's Law, the ray emerges at 54.3 degrees.

35. Remember to find the angle from the normal! Theta 1 = 64.3 degrees, n1 = 1, n2 = 1.33, so theta 2 = 42.6 degrees. 2.1 meters * tan (42.6 degrees) = 1.93 meters. The total distance from the wall is therefore 2.7 m + 1.93 meters = 4.63 meters.

Can YOU tell me what's on the other side of the mirror?

Hi everyone,

In case you had trouble watching the lesson in class Thursday and Friday, here is the site:

http://www.hippocampus.org/AP%20Physics%20B%20II

Click on 'Flat and Concave mirrors" under 'Geometric Optics' in the menu. You can also watch the Refraction video there as well.

Don't forget - we will have a ray tracing quiz and a problem from the handout. Feel free to check with me or each other to make sure you agree on the answers. Thanks to all of you who have already done this!

I'm Talking 'Bout the Robot In the Mirror...

Q19. Since the image is clearly magnified, this must be a concave mirror. Only concave mirrors can magnify objects. This occurs when an object is placed between the mirror and its focus.

P4. The easiest way to solve this problem is to extend the top ray through the mirror. You then obtain a pair of similar triangles. Solving the proportion, x = 75.9 cm

P11. A picture is worth a thousand words:

P16. Leave out part b, we will discuss the equations of ray tracing tomorrow in class. Make a scale sketch of the situation on graph paper (notice it is a convex mirror!) - you should obtain that the distance of the image is between 6 and 7 centimeters on the far side of the mirror, and the image should be upright.

Beats, Electromagnetic Waves, and something else for dessert....

First, some web-based toys to simulate beats and other bits of interference:

http://www.mta.ca/faculty/science/physics/suren/Beats/Beats.html
See two sine waves added together.

http://micro.magnet.fsu.edu/primer/java/interference/waveinteractions2/index.html
Phase Difference Demo

http://library.thinkquest.org/19537/java/Beats.html
Simulation of beats that you can hear!

Second, the table of data you must use for calculating the speed of sound from today's experiment:


Third: Some solutions to the problems from today's homework set:

Chapter 12:
44. The unknown whistle has to be HIGHER than the 23.5 kHz if it is to be inaudible to humans. Humans can hear up to around 20 kHz. This means the whistle frequency must be 23.5 Khz + 5 kHz = 28.5 kHz.

53. 77.27 Hz is the frequency observed by the stationary tuba player - it is higher by the Doppler effect. The beat frequency would be 2.27 Hz.

54. This problem can be figured out in much the same way as 53, just with the actual frequency of the horn as the unknown. If the stationary frequency is F_o, the higher frequency perceived by the stationary observer must be (340/325)*F_o.

The equation (340/325)*F_o - F_o = 5.5 Hz will allow you to determine the actual frequency of 120 Hz. The two frequencies would be 120 Hz and 126 Hz, the 126 Hz observed from the moving car. Thus the frequency of both horns must be 120 Hz.

Your interference doesn't PHASE me...

I apologize for this, but the bulk of the homework I assigned today has to do with beats, which I left out in order to talk about interference. I will be sure to give you some interference problems over the next few days to compensate.

Tomorrow's lab will hopefully take the first period, and then we can talk about beats and electromagnetic waves during 7th.

Question 17.
You can use either the equation we derived at the end of class, or reasoning about the distance traveled by each wave. If the sources are close together, then the distance traveled by each wave will be similar, and the waves will have to travel a larger difference in distance in order to reach the path difference needed for constructive or destructive interference. This means the locations will be further apart when the sources are close together.

Introducing the Doppler 5 x 105...

Hi everyone,

Please make sure you bring in the waiver form tomorrow - if you need to download a new copy, please click here.

Solutions:

Question 17: The relative speed between the whistle and the listener is highest at point C. By today's discussion of the Doppler effect, this must also be the location of the highest frequency.

51.
a) Vobs = 370 m/s, Vsource = 340 m/s, fobs = 1959 Hz
b) Vobs = 310 m/s, Vsource = 340 m/s, fobs = 1631 Hz

52. The frequency of the wave as heard by the moving object is 46323 Hz. This frequency now acts as the source, and the bat is the observer. The new frequency heard by the bat is 43150 Hz.

78. You should be able to set up two equations with Fsource and v(the speed of the train) as the unknowns. Let Vs stand for the speed of sound.
First equation: Vs/(Vs - v) = 522/Fsource
Second equation: Vs/(Vs + v) = 486/Fsource.

Dividing the equations, Fsource drops out, and you can solve numerically to get v = 12.14 m/s.


Sites for Doppler Effect:

http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=21.0

Resonance in the Modern World

First, some interesting sites related to today's discussion:

http://library.thinkquest.org/19537/cgi-bin/showharm.cgi
This site contains a program that allows you to put together harmonics and see what they sound like together.

http://www.pbs.org/wgbh/nova/bridge/meetsusp.html#clips
This site describes one of the most catastrophic applications of resonance, the Tacoma Narrows Bridge collapse.

And here is a video of Tuvan throat singers, as promised.

Now, some solutions:

33. Since the nodes are now at half the distance, and the wave velocity is the same, the frequency must double to 8 Hz. This is beyond the highest frequency of earthquakes.

36. The tube must be closed at one end - look at the ratio of the frequencies to each other, as they are odd fractions. The fundamental frequency is 88 Hz.

37. f = 5v/2L, so v = 247.5 m/s.

38. Open at both ends, 7v/2L = 33o Hz, L = 3.606 m

75. n*lambda/4 = L1, (n+2)*lambda/4 = L2. Subtract the two equations, and you get that lambda = 2(L2 - L1), leaving that f = 630 Hz if the speed of sound is 340 m/s.

You're Standing in my Wave....

As I mentioned in class today, I expect you to be checking this blog on a daily basis to get answers for the HW assignments. This is more of an issue for those of you that are NOT staying for tutoring during 8th or after school.

below you will find a list of the last four digits of the OSIS numbers of people in the class, along with a five letter code word. Please post a comment or send me an email with your codeword to confirm you have checked the blog.

8950 - bbjah
1893 - qpkaj
3853 - zzytn
2615 - jjajs
9409 - quahl
6872 - hhsga
9021 - bnxma
6990 - jqpwo
6352 - nnzna
2531 - mqpao
1679 - mansp
9200 - gsgah
9226 - bbzna
0598 - nnqma
6706 - mpaoq


Q25. The nodes are locations of zero amplitude, so these positions could be touched without disturbing the wave.

Problems:
18. amplitude = 0.3500 meters, frequency = .8754 Hz, period = 1.142 seconds, total energy = .7411 J, KE = .6807 J and PE = .0604 J

51. 440 Hz, 880 Hz, 1320 Hz, 1760 Hz (f_n = nv/2L)
53. 70 Hz, 140 Hz, 210 Hz, 350 Hz
55. nodes are always a half wavelength apart from each other - wavelength is 19.37 cm, so nodes are half of this, 9.685 cm.
56. 87.5 Hz, n = 3 and 4 for the given frequencies.
60. m = 4f²L²*mu/(n²g), so m = 1.421 kg, .355 kg, .0568 kg for (a), (b), and (c).

Web sites from today:

http://www.colorado.edu/physics/2000/applets/fourier.html- Constructive/Destructive Interference

http://home.austin.rr.com/jmjensen/JeffString.html

Reflected Waves


http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=19 - Waves traveling in opposite directions making standing waves



Standing Wave - nodes and antinodes

http://id.mind.net/~zona/mstm/physics/waves/standingWaves/standingWaves1/StandingWaves1.html - Vibrating string, standing waves fixed at both ends.

http://www.walter-fendt.de/ph11e/stlwaves.htm - open/closed Tube with standing waves

Wave - You're on National Television!

For those of you on the trip today:

The homework is to read P. 322 to 328, and DO Q 10, 18, P 34, 35, 39, 41, 50

We will be having an AP problem wednesday quiz - one EMI problem, another on momentum.

Here are the quiz solutions.

Q10 . The water sloshing back and forth has a specific frequency associated with it because of the size of the pan and the speed of the waves on the surface of the water. The only way to get the water to slosh is if the swinging occurs at this special frequency.

Q18 Hitting across the end creates a transverse wave, hitting perpendicular to the end creates a longitudinal wave.

Problems.
34. 2.83 m/s
35. 1.259 m
39. 0.332 s
41. The speed of sound in water is 1440 m/s. The sound must travel down to the ocean floor and back. The distance is therefore 2160 meters.
50. The string may be flat, but it is not at rest - try one of the sites below to see what happens when two reversed pulses pass each other on a string.

EMI is good for the teeth and bones.

1986B4.
a) 3 V
b) clockwise
c) 0.6 N
d) 1.8 W

1982B5.
a) 0.06 Webers
b) 0.06 V
c) -0.3 A from t = 0 to t = 2 seconds, 0 A from t = 2 to t = 4 seconds, and +0.15 A from t = 4 to t = 6 seconds.

3.
a) final speed is (2gy0)^(1/2)
b) I = Bh/R*(2gy0)^(1/2)
c) flux is increasing into the page, so induced current increases the flux out of the page. Current is CCW.
d) For flux:
Flux increases linearly to whB at w; Stays constant until 3w, and then decreases to zero at 4w.
For current: The magnitude of I is the value given in part (b).
Positive I from 0 to w, 0 from w to 3w, -I from 3w to 4w, and then zero from 4w to 5w.

4.
a) m = ILB/g = B^2L^2v/gR = 0.143 kg
b) 1.51 J
c) 1.51 J

5.
a) 0.3 Volts
b) 0.06 A, counter clockwise
c) 0.018 W
d) F = 0.036 N
e) 20 more turns means potential difference increases by a factor of 20. This also means twice the resistance of the wire. Since current I = deltaV/R, the factors of 20 will divide out, leaving the current the SAME.

Some toys for Tuesday's lesson:

http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=18 pulse superposition

http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=323.0

http://www.colorado.edu/physics/2000/applets/fourier.html- Constructive/Destructive Interference

http://www.surendranath.org/Applets/Waves/TwaveRefTran/TwaveRefTranApplet.html - simple reflection

EMI solutions and a form!

Hi guys,

Thanks to all who came to robotics this weekend - it was quite a blast. I hope the REACH session was also useful for those that went. Put the next one on your calendar, March 28th at Baruch College.

Please complete the form below to let me know when you will be meeting with me.



Solutions:

I apologize for leaving it out, but please use the image below with Problem 2 from the handout from today's lesson:





Questions:
15. The change in flux causes an induced current within the aluminum. The magnetic field created by the induced current is attracted to the magnetic field of the bar magnet, opposing the force tending to pull the sheet out of the magnet. This induced magnetic field forms because of the induced current, not because the magnetic properties of aluminum.

16. We discussed this in class!

Problems:
3. Counter clock-wise

4. Field lines going out from magnet to the right. Moving it through loop increases the flux going into loop towards the right. By Lenz' law, the induced current will increase the magnetic flux going into the loop towards the left. This will create a current that flows clockwise through the loop, and from right to left through the resistor.

7.
a)The current through the outer loop decreases since the resistance increases. This means that the magnetic field (and therefore the flux) through the inner loop out of the page is decreasing. By Lenz's law, the induced current will attempt to increase the flux out of the page to oppose the change. This will come from a counter-clockwise current.
b) If the loop was moved outside to the left, the flux would be decreasing into the page through the loop. The induced current would try to increase flux into the page through the loop, which would result from a clockwise current.

8.
a) counter-clockwise
b) clockwise
c) zero (no change in flux)
d) counter-clockwise

EMI Mr. Weinberg, I DID need to know that....

Question 2:
Magnetic field describes how the influence of magnetic force is transmitted through space. Magnetic force describes how magnetic flux passes through a specific region of space, specifically a loop or area through which magnetic field lines travel. It is possible to have magnetic field and not have magnetic flux (if area = 0 or if the angle between B and the normal vector is 90 degrees), but it is not possible to have magnetic flux without magnetic field.

Problems:
2. 0.147 Volts

10.
a) 0.169 Volts

12.
b) 4.29 x 10^-2 Volts
c) 17.2 mA

13.
a) .841 m/s
b) .757 N/C

89. 7.27 x 10^-3 J

Here are a couple of sites with fun EMI tools to play with:

http://www.ngsir.netfirms.com/englishhtm/Induction.htm
http://higheredbcs.wiley.com/legacy/college/halliday/0471320005/simulations6e/index.htm?newwindow=true

Field, Force, Wire, Charge - Performing This Thursday at 12:25 PM

Hmm...tomorrow is Wednesday, isn't it?

I think it's time for an AP question...think Thermo.


1.
a) 5 x 10^-5T
b) West
c) Net B-field is of magnitude 7.07 x 10^-5T directed at 45 degrees North of West. (It is not enough to say 7.07 x 10^-5T at 45 degrees!)

2. a = 0.0076 m/s^2

3. Since the force is attractive between the two wires, the current must be going in the same direction. k'I1I2/d = 6 x 10^-4 N/m...solving for the unknown current gives I = 48 A



4.
a) Remember that force/unit length means that from F = IBL, force per length is F/L = IB. Thus F/L = 2.5 A * 1.10 T * sin(90) = 2.75 N/m.

b) F/L = 2.5 A * 1.10 T * sin(45) = 1.944 N/m

5. F = IBL so I = F/(BL) = 11.67 A

6. Let's define a magnetic field OUT of the page to be positive.

a) B_total = kI1/d = (2 x 10^-7 T*m/A*10 A/.05 m = 4 x 10^-5T

b) B_total = k'I1/d + k'I2/d = (2 x 10^-7 T*m/A*10 A/.05 m + (2 x 10^-7 T*m/A*5 A/.05 m= 6 x 10^-5T

c) B_total = k'I1/d - k'I2/d = (2 x 10^-7 T*m/A*10 A/.05 m - (2 x 10^-7 T*m/A*5 A/.05 m= 2 x 10^-5T

7. F = IBL sin(60) = 1.212 N

8. This is an AP problem - don't be scared off by it! Remember to set a direction (in or out of the page) as positive and then sum all fields together, being sure to set fields positive or negative depending on the direction of each. We will use out of the page as positive.

Sum of fields at P:
-k'(1A)/(0.5 m) + k'(3A)/(1.0 m) = 2 x 10^-7 T

This shows that the field is directed out of the page (a) at the given magnitude of 2 x 10^-7 T.

c) Force is to the left, B is out of the page, and v is towards the top of the page. This only matches the left hand rule, which means the charge must be negative.

d) Fmag = qvB sin(90) so q = Fmag/(vB) = 5 x 10^-7 C

e) An electric force to the right of magnitude 10^-7 N would make the net force zero. Since Felec = qE, the field would have magnitude 0.2 N/C. Since the charge is negative, the electric field would need to be directed to the left for the force to be to the right.

Wire You Forcing Me To Stay Current?

Q11. Since the North pole is on the left, and the South pole is on the right, the B field lines go towards the right. Since the current is going into the page, and B is going to the right, by the RHR the force will be towards the top of the page.

Q17. The kinetic energy will stay constant since the magnetic force (which is also the net force acting on the particle) is always perpendicular to the velocity. This means acceleration is always perpendicular to velocity, which only occurs when an object is in uniform circular motion. Uniform circular motion is characterized by a constant speed, and by KE = 1/2mv^2, a constant kinetic energy.

17.You can find the acceleration first using kinematics, since you know displacement (1 m), final velocity (30 m/s), and initial velocity of zero since it starts at rest.

a = 450 m/s^2

Using Newton's 2nd and a FBD, you can also find that the magnetic force is the only force acting to accelerate the bar.

Fnetx = IBL = ma so I = ma/BL

I = 1.985 Amperes, magnetic field is directed into the page through the loop formed by the bars and the rails.

66.
a)
Remember that old thing called the Free Body Diagram?

We need to use it here - the magnetic force on the current accelerates the sliding wire down the rails.

Fnetx = Fmag = IBl = ma so a = IBl/m.

We can use v = vo + at to write the velocity as a function of time. The rod starts at rest, so vo = 0.

Thus v(t) = IBLt/m

b) With friction acting, the y-direction (which is directed out of the page in the photo above) becomes more important.

Fnety = Fn - mg = 0 so Fn = mg.

Since friction force Fk = mu*Fn, Fk = mu*mg

The new Fnetx = Fmag - Fk = ma
IBl - mu*mg = ma
a = (IBL/m) - mu*g

Using v = vo + at again, v(t) = 0 + ((IBL/m) - mu*g))t

c) Using the right hand rule, we can point our fingers North, and palm out of the page to go along with B. This leaves our thumb pointing to the right. This is then the direction of force.

Try it - don't guess!

"I'm Sorry, I'll Need to Search Your Bag"...and other funny things said after Mass Spectroscopy....

Q12. Magnetic force requires a charged particle to have velocity. Thus the magnetic force will always be zero unless the particle starts with some velocity. An electric field exerts a force on any charged particle of magnitude F = qE, and therefore, can make an electron at rest accelerate.

P5. Using the left hand rule - the force is directed South. The magnitude will be 7.45 x 10^-13 N

P9. To produce a circular path, the magnetic field is perpendicular to the velocity. The magnetic force
provides the centripetal acceleration:
qvB = mv2/r, or
r = mv/qB;
0.25 m = (6.6 ´ 10–27 kg)(1.6 ´ 107 m/s)/2(1.60 ´ 10–19 C) B, which gives B = 1.3 T.

P14. The equation we derived yesterday comes up in this problem: qvB = mv^2/R. The difficult part is that we can write speed in terms of the radius: 2*pi*R/T = v. Making this substitution results in the expression .5mv^2 = 2*pi*qBR^2/T where T is the period of revolution.

17. skip this - we will do this in class Friday.

55. This is very similar to our result from class, but we don't know the speed of the particle when it enters the magnetic field. We DO know the particle is accelerated through a potential difference V...How can we relate a change in potential of a charge Q to the final speed it has if it starts at rest?

How Could She Not Like you? You have a Magnetic Personality!

Hi everyone,

1. As a recommendation, you might want to check out questions 7 and 8 for more practice with the right/left hand rule. I'll post the solutions below.


As promised, here is a video of a magnetically levitated frog.


You can search on Google for more interesting videos if you enter 'magnetic levitation'.

The magnetic force simulator is located at
http://www.physics.upenn.edu/courses/gladney/phys151/lectures/lecture_feb_24_2003.shtml

You must stop and reset the simulation each time you change something to see the effect of the change.

Here are the HW solutions from the handout:
1. 3.84 x 10^-14N
2. into page, towards left, towards top of page, towards right, force is zero!
3. For this problem, you have to say which direction you are assuming each vector is pointing towards. If velocity is towards the right side of the page, and B is 30 degrees above the velocity towards the top of the page, then you must use the perpendicular component of one vector along another and the right hand rule. This results in a 3.2 x 10^-16N force directed out of the page for (a).
b) 1.92 x 10¹¹m/s^2
c) The force has the same magnitude as (a) but the force and acceleration are directed into the page. The acceleration of the electron would be 3.51 x 10¹¹m/s^2
4. 9.375 T towards top of page

Solutions to 7,8:


7. To find the direction of the force on the electron, we point our fingers in the direction of v and curl them
into the magnetic field B. Our thumb points in the direction of the force on a positive charge. Thus the
force on the electron is opposite to our thumb.
(a) Fingers out, curl down, thumb right, force left.
(b) Fingers down, curl back, thumb right, force left.
(c) Fingers in, curl right, thumb down, force up.
(d) Fingers right, curl up, thumb out, force in.
(e) Fingers left, but cannot curl into B, so force is zero.
(f) Fingers left, curl out, thumb up, force down.

8. We assume that we want the direction of B that produces the maximum force, i. e., perpendicular to v.
Because the charge is positive, we point our thumb in the direction of F and our fingers in the direction
of v. To find the direction of B, we note which way we should curl our fingers, which will be the direction
of the magnetic field B.
(a) Thumb out, fingers left, curl down.
(b) Thumb up, fingers right, curl in.
(c) Thumb down, fingers in, curl right.

From the handout used on Thursday:

1.
a) 8kQ^2/(5^3/2)a^2
b)E = kQ/4a^2 directed to the right
c) -9kQ/2a
d)very similar to the graph of y = x*e^(-x^2)
e)v = (9kQq/m)^(1/2)

2.
a) left (-), right (+) from E field lines
b) 100 V
c) 1.3 x 10^-10F
d) 8.0 x 10^-16N directed to the right
e) by conservation of energy, v = 4.2 x 10⁶m/s

3.
#1.
i. 480 ohms, .25 A
ii. 360 ohms, .33 A
#2.
The resistances are the same as above. Since they are in series, the current rhough both is 0.143 A.
#3. in order, moving down the list: 2, 1, 3, 4
d) parallel: 70 W, series, 17.2 W

4.
a) 20 V
b) Q = 3.0 x 10^-8C
c)
i. 30 V since current is zero!
ii. E = 0 inside any conductor.
iii. With 30 V over the two gaps, using V = Ed, E = 60,000 N/C

5.
a) t = L/v₀
b) a = Dv₀²/L²
c) E = mDv₀²/qL²
d) V = mD²v₀²/qL²

Electric Current exam tomorrow. You might also plan to meet with your partner to get an idea of what you will be trying during the break for the battery project.

RC Cola Isn't Just A Soft Drink....

3.
a) 12-kOhm: .333 mA, 15 kOhm: .333 mA, 3 kOhm: 0 A
b) 50 microCoulombs

4. 2A

5.
a) 0.4 seconds
b) zero (capacitor and resistor are in series!)
c) 60 microCoulombs

6. 72 microCoulombs

From the book:
51.
a) 4.129 x 10-5C
b) through .5, 6, and 5 ohm resistors: I = .635 A, through 8 ohm = .212 A, through 4 ohm = .424 A.

52.
a) C = 2.33 x 10-9F

54.
a) 8V
b) 16 V
c) 8 V
d) 5.76 microcoulombs

Capacitors - the NEW resistors? More on Page 7!

40. minimum C = 1.36 x 10-9F connected in series, maximum C = 1.95 x 10-8F connected in parallel.

41. 300 pF connected in parallel.

42. C1 + (C2C3/(C2+C3)), Q on C1 = 562.5 microcoulombs, Q on C2 = Q on C3 = 375 microcoulombs

44
(a) For 0.4 microfarad capacitor, Q = 2 microcoulombs, V = 5 V
For 0.5 microfarad capacitor, Q = 2 microcoulombs, V = 4 V
(b) Both capacitors have 9V across them, For 0.4 microfarad capacitor, Q = 3.6 microcoulombs, 0.5 microfarad capacitor, Q = 4.5 microcoulombs

72. (a) When the capacitors are connected in parallel, we find the equivalent capacitance from
Cparallel = C1 + C2 = 0.40 µF + 0.60 µF = 1.00 µF.
The stored energy is
Uparallel = 0.5*Cparallel*V^2 = (1.00 x 10–6 F)(45 V)^2 = 1.0 x 10^–3 J.
(b) When the capacitors are connected in series, we find the equivalent capacitance from
1/Cseries = (1/C1) + (1/C2) = [1/(0.40 µF)] + [1/(0.60 µF)], which gives Cseries = 0.24 µF.
The stored energy is
Useries = 0.5*CseriesV^2 = 0.5*(0.24 x 10–6 F)(45 V)^2 = 2.4 ´ 10^–4 J.
(c) We find the charges from
Q = CeqV;
Qparallel = CparallelV = (1.00 µF)(45 V) = 45 µC.
Qseries = CseriesV = (0.24 µF)(45 V) = 11 µC.

There was some internal resistance to the revolution - Batteries & Internal Resistance

2.
a) 16 ohm current = .106 mA
b) battery current = 6.37 mA
c) New Req = 1413.8 ohms, terminal voltage = 8.987 V

3.
a) 2 A
b) 8V

From Textbook:
18 a) 8.406 V, b) 8.491 V
20. r = .4068 ohms
21. .06 ohms
29.
For the conservation of current at point c, we have
Iin = Iout ;
I1 = I2 + I3 .
When we add the internal resistance terms for the two
loops indicated on the diagram, we have
loop 1: V1 – I2R2 – I1R1 – I1r1 = 0;
+ 9.0 V – I2(15 W) – I1(22 W) – I1(1.2 W) = 0;
loop 2: V3 + I2R2 + I3r3 = 0;
+ 6.0 V + I2(15 W) + I2(1.2 W) = 0.
When we solve these equations, we get
I1 = 0.60 A, I2 = – 0.33 A, I3 = 0.93 A.

Oh, and here's something interesting I found:

February break clue #2:

February Break Assignment - Clue #1

Clue #1

Solutions will be posted tomorrow. Here are the solutions to Friday's work:

Problem 3.
a) This is a standard problem where you have to find Req. The trick here is to realize that the 3 ohm and 6 ohm resistors are in parallel since their ends are connected directly together. You should obtain that Req = 6 ohms for the entire circuit. The battery current is therefore 1 ampere, and since the 4 ohm resistor is in series with the battery, it also has a current of 1 ampere.

b) To show that the junction rule holds, you must demonstrate that the current into the junction (which is the current through the 3 ohm and 6 ohm resistors) is equal to the current out (the current going through the 4 ohm resistor). This emans the only thing missing is the current through the 3 ohm and 6 ohm resistors.

You can use Ohm's law to find that the voltage drop across the 4 ohm resistor is (4 ohm)*(1 A) = 4 V, which means the voltage left to drop across the 3 ohm and 6 in parallel is 2 V. Since both resistors are in parallel, they BOTH must have 2 volts across them. This allows you to calculate the current through them by Ohm's law, 2/3 A and 1/3 A respectively. Since 2/3 A + 1/3 A (current in) = 1 A (current out) we show that the node rule holds.

c) The only thing you must show is that for the large loop, the sum of potential differences adds to zero. Going clockwise from the negative terminal of the battery:

6 V - (2/3 A)(3 ohms) - (1 A)(4 ohms) = 0.

The rule holds!


1989B3.

a)
i. 40 W (direct application of P = IV)
ii. 20 W (calculation of power for a constant force, P = F*v)
iii. 60 W (the battery is supplying ALL of the power to the circuit, so it must be equal to the sum of the power used by the other circuit elements.)

b)
i. 20 V (Ohm's law)
ii. 10 V (You know the power is 20 W, and the current through the motor is 2 A. Using P = IV, the voltage must be 10 V)
iii. 30 V (Similar to part (a), the battery is the source of all potential difference in the circuit, so it must be the sum of the potential differences for the motor and resistor.)


c) 15 V (You can get this by calculating the new power required to lift the mass and using P = IV, or by using the fact that it says the voltage is directly proportional to the speed of the motor. )

d) 7.5 ohms (15 V potential difference across the motor, and the battery potential difference is still 30 V. Thus the resistor has 15 V across it, with the same 2 A current. This makes the resistance 7.5 ohms by Ohm's law.)

1983B3

a) 5 ohms
b) i. 4/3 A ii. 2/3 A
c) At point B: 10 V, at point C: -10 V, at point D: -2V
d) 40 W

1982B4

a) clock in parallel with the battery, radio in series with a resistor, and together in parallel with the battery.
b) 600 ohms
c) P = .45 W, energy = 27 J for a minute

Also, here are a couple more circuit problems to make sure you're where you need to be. The answers are included. Click to see them full size (unless your eyes are really THAT good....)

Dr. StrangeCircuit, or How I Learned To Stop Worrying And Love Finding Req.

59. 99.8 Ohms
60. 6.762 Ohms
61. 4.601 Ohms
62. I = .083 A, P = .833 W
63. P = 2.222 W
64. R = 24.94 Ohms
65. current through 20 Ohm resistor = .75 A, current through 9 Ohm resistor = 2.11 A

Series? Parallel? Neither? This question offends me.

Equivalent Resistance HW:
1.
a) 2/3 R
b) 2 R
c) 15/2 R

2.
a) Req = 6 Ohms
b) Req = 5 Ohms

Book HW:

Questions:
2. Parallel lights - more difficult to connect, but all lights will stay on even if one goes out.
series lights - easy to connect lights together, but if one goes out, they all do!
4. For the bulbs in series, the current is the same. By P = I²R, the bulb with the largest resistance will have the highest power usage, and will therefore be the brightest.

If the bulbs are connected in parallel, the voltage is the same for each. The power equation becomes P = V²/R, which means the larger power will be used by the bulb with less resistance.

Problems.
2.
a) 360 Ohms
b) 8.89 Ohms
4. Maximum 2800 Ohms, minimum 261.4 Ohms
7. 4550 Ohms
13. 105.2 Ohms
24. I = 0.409 A.

MC Solutions and Results

Hi everyone,

Here is a PDF of the multiple choice solutions and the class results sorted from the worst to the best questions for our class.

Please go through and start looking at the questions so you come in Tuesday with some ideas for what did/didn't go well.

See you all Tuesday,

EMW

Midyear Review Solutions

The solutions are posted below - I have updated some of the previous answers with a bit more details, so be sure to check here for the latest answers.

Please also post a comment to this blog post on what topics/problem types you personally want the most help with during class tomorrow.

These are the problems I have created for tomorrow - I suggest you try them and come in with questions: Review problems for Monday

Midyear Review #1 - 1D kinematics
1.
(b) slope is zero, zero acceleration
(c) zero
(d)&(e) 200 meters
2.
(b) 8 m/s^2
(c) area under v vs. t graph, distance = 100 m.
(d) Y = vo*t + .5at^2 = 100 m.
(e) 16 m/s
(f) graph should be a parabola with its vertex at t = 0, y = 0.

3.
(a) 4 seconds
(b) horizontal line from t = 0 to t = 3, then a diagonal line going down to zero.
(c) 60 meters
(d) 100 meters

4. 24 m/s

5.
(a) +15 m/s, -10 m/s
(b) change in velocity is -25 m/s (or 25 m/s directed upwards)
(c) 50 m/s^2 directed upwards

6.
(a) 4 m/s^2
(b)East
(c) velocity directed south, acceleration directed East

Midyear Review #2 - 2D kinematics and Projectile Motion:
1.
(a) 1.563 seconds
(b) 7.813 meters
(c) 13.53 meters

2.
(a) 4.0 seconds
(b) 5 m/s
(c) (We can assume that the net force is mostly vertical since the horizontal velocity (5 m/s) is so much less than the vertical velocity (40 m/s). acceleration = .769 m/s^2, displacement = 1040 meters (that's a deep target!)

3.
(a) 412.3 km at 14 degrees North of East.
(b) 42.7 km/h
(c) 51.7 km/h

Midyear Review #3 - Newton's Laws


1. 4 N directed upwards

2. T1 (diagonal cord) = 200 N, T2 (horizontal cord) = 173 N

3.
(a) a = F/6m
(b) Block 1 (3m), Block 2 (m), Block 3 (2m)
Fnet1 = F/2
Fnet2 = F/6
Fnet3 = F/3
(c) T1 (between 3m and m) = F/2, T2 (between m and 2m) = 2/3F


4. The key thing to think about is that WALL-E always hangs from EVE's arms. The FBD of WALL-E always has EVE's force directed upwards, and his weight directed downwards. The difference is the direction of acceleration, which must always be directed towards the center of the circle.

For (a), the acceleration is upwards, so F – mg = mv^2/R. F = 669 N.

For (b), the acceleration is downwards, so mg – F = mv^2/R. F = 81 N.

5.
(a) T1 = 60 N, T = 100 N, minimum value of static friction coefficient = .27. Notice that this is the solution if you say that static friction is directed DOWN the inline.

If you instead make the static friction force go UP the incline in your FBD, you get a negative value for mu, which means that this is NOT possible in this situation.

(b) a = 1.31 0.670 m/s^2, T = 84.9 91.3 N. Notice that it says to assume the incline is smooth, which means no kinetic friction.

c) If the cord is cut, the right block will fall at g, or 9.8 m/s^2. The left block will slide at g sin(60) = 8.49 m/s^2. The answers must be different because the two blocks are no longer linked by the cord.

Midyear Review #4 – Circular Motion, Gravity, SHM

1. Be sure to use the static friction coefficient, as static friction is what keeps the car from slipping off the road. Kinetic friction occurs when two surfaces slide past each other.

Max speed = (mu*g*r)^(1/2) = 23.7 m/s

2.
(a) Mass is the same, 70 kg
(b) 700 N (16) = 11,200 N
(c)160 m/s^2


3. Remember to draw a FBD of the satellite, with the gravity force directed towards the Earth's center, and positive also directed towards the center. You should get that v = (GM/r)^(1/2).

4.use your answer to #3 to justify your answer here. Also, notice that the mass does NOT change your answer. The ratio is equal to (1/5)^(1/2)

5.
a) Since the mass is not moving when the spring is at maximum compression, the KE = 0.

b) U = 1/2*k*delta_x^2, so the potential energy = 1/9*maximum U = 8 J.

c) Since the total energy is constant (energy is conserved!), the KE = 64 J.

d) T = 2*pi(m/k)^(1/2), but since you aren't given the mass in this problem, you can't solve it.

MARLON: You're slacking!

I know, I know. So let's suppose the mass was 1 kg. This would mean the period would be 0.363 seconds.

e) If the pendulum has the same frequency as the mass on the spring, it must also have the same period. The period of oscillation for a pendulum is given by T = 2*pi(l/g)^(1/2) so for a period of 0.363 seconds, this means the length would have to be 3.27 cm. That's a small pendulum, so it will swing back and forth fairly quickly.

MARLON: That pendulum is NOT slacking!

I couldn't agree with you more.

6.
speed, max D, min A
kinetic energy max D, min A
potential energy, max A, min D
total energy, constant
angular momentum, constant
net force on the satellite, max D, min A
acceleration, max D, min A

Capacitors? Yes you CAN!

Reminder: The quizam will have a mixed bag of electrostatics AP problems, and might have another easy-to-predict surprise....

here's a hint:

12:41 PM, January 20, 2009

Solutions: Chapter 17, Q13, P32, 34, 37, 39, 47

Questions:

Problems:
32. Q = CV, so V = 22.0 volts.

34. using the equation for a parallel plate capacitor: C = e₀*A/d so the area = 5.0 x 10⁷ m^2. That's a lot of area!

37. We can get the electric field using V = Ed. Also, Q = CV.

We also need the fact that for a parallel plate capacitor, C = e₀*A/d.

We can put ALL of this together to get that Q = e₀*A*E*d/d = e₀*A*E , all of which we know! Substituting, Q = 2.63 x 10^-8 C.

39. Q = CV, so V = 90 volts. The potential difference V = Ed, so E = V/d = 4.5 x 10⁴ V/m or N/C.

47. The key to these problems is to use the equation in the form that does NOT change. For part (a), for example, it does NOT make sense to use U = 1/2*QV because the charge Q depends on the voltage through Q = CV. If you instead use U = 1/2CV², C is a constant, and V is the only thing that changes.

a) The energy must double, since U = 1/2CV² and C is a constant.

b) In this case, the Q is changed, and C must stay constant. We can then see by U = Q²/2C that the energy must be multiplied by a factor of 4.

c) If the battery stays connected, it is the VOLTAGE that stays the same, but the capacitance is divided by 2. Since U = 1/2CV^2, the energy will be divided by 2.

How much for a charged particle in mint condition?

Definition quiz Friday. Word.

Click here for the solutions to the problems from class.

From Infinity...to a distance R!

For those of you reading the blog, just a heads up that tomorrow IS Wednesday, our day for an AP problem quiz. Another bit of useful info...it will HEAVILY be based on kinematics. Might be a good idea to check out those review problems on kinematics handed out in class - the solutions are posted on the blog in January 2008. Use the menu on the right hand side to find those solutions....

Now for the solutions:

p. 522, 11, 12, 15, 16, 18, 20, 23

11. KE = 1/2*mv^2 with KE = 28 x 10⁶ eV * (1.6 x 10^-19 J/eV) = 4.5 x 10^-12 J. v = (2*KE/m)^(1/2) = 7.3 x 10⁷ m/s.

12. The same procedure, except we use the mass of the alpha particle. v = 1.6 x 10⁷ m/s.

15. We must first find the total potential at each location. The first location has a distance of 16 cm from both charges. Using V = kQ/R and adding the potential for each charge, the total V in the middle is 3.38 x 10⁶ V.

In the second position, the distance from one charge is 6 cm, and the other is 26 cm. We again use V = kQ/R and add the two results together to get 5.54 x 10⁶ V.

We can neglect the change of kinetic energy during this process because the charge is being 'placed' at each location.

Wncf = delta_K + delta_U = q*delta_V = 1.08 J

16.
a) The electric potential of the proton is V = ke/R = 5.8 x 10⁵ V

b) The potential energy of the system is found by multiplying the potential created by the proton (the answer from a) by the proton located at the given distance. Since V = U/q, U = q*V = 9.2 x 10^-12 J.

18. This follows the same pattern as what we did at the end of today.

The first electron takes zero work to bring together.

The second electron takes e*delta_V = -e*(k(-e)/R - 0) = ke^2/R.

The third electron must overcome the repulsive forces of BOTH electrons that are already there: e*delta_V = -e*(k(-e)/R - 0) - e*(k(-e)/R - 0)= 2ke^2/R.

The total must then be the sum of these quantities, 3*ke^2/R = 6.9 x 10^-18 J.

20. Energy is conserved, so:
initial PE + initial KE = final PE + final KE
ke*Q/R + 0 = 0 (since it is at infinity) + 1/2*mv^2
v = 2.33 x 10⁷ m/s.

23. We first need to find the total potential at each location.

At A:
V_A = kq/b + k*(-q)/(d - b) which can be simplified to kq(d - 2b)/b(d-b).

At B:
V_B = kq/(d - b) + k*(-q)/b = kq(2b - d)/b(d - b).

The difference in potential Vb - Va can be simplified to 2kq(2b - d)/b(d-b).

You have a lot of POTENTIAL...but you need to WORK harder.

p. 522, Q7, P 4,5,6,8,9

Question 7. Electric potential is the amount of energy per unit charge, while electric field is the amount of electric FORCE per unit charge. Electric potential is just electric potential energy divided by charge.

Problems.
4. If the electron gains 3.6 x 10^-16 J of KE, it means it loses a potential energy of 3.6 x 10^-16 J. This happens as it crosses a potential difference delta_V = delta_U/q.

Thus delta_V = (3.6 x 10^-16 J)/(-1.6 x 10^-19 C) = 2,156 V.

Since it is an electron with a negative charge, the final plate (plate B) is at the higher potential.

5. delta_V = Ed, E = delta_V/d = 220 V/(.0052 m) = 4.23 x 10⁴ N/C or V/m.

6. Following the same procedure as with #5, the answer is 7.04 V.

8. d = delta_V / E = 3.3 x 10^-5 m

9. Since we have a non-conservative force doing work, a change in kinetic energy, and potential difference, it sounds like we might be able to use the generalized work energy theorem.

W_ncf = delta_K + delta_U
25 x 10^-4 J = 1.82 x 10^-4 J + delta_U
delta_U = 2.02 x 10^-3 J
delta_V = delta_U/q = 2.02 x 10^-3 J/(-7.5 x 10^-6 C = -269 V. This is the potential difference from a to b for the negative charge. The potential difference Va - Vb (the reverse of Vb - Va) = +269 V.

Charging by Induction, Conduction, Frustration

HW: Q7,8, P19, 27, 45, 56
Questions:

7. The attraction of the negative charges to the positive charges is strong, but remember that the charges will also feel a repulsion force from being close to the other negative charges.

8. The leaves of the electroscope act like springs as they are forced apart. You can feel this force in a simple experiment - take a ruler and hold it against the edge of a table so that most of its length is hanging off the edge. As you push on the free end of the ruler, you will feel a force that is proportional to how far you push on the end (displacement).

Problems:
19. To figure out where the third charge must go, you have to observe that it can't be located between the charges, because the forces from each charge would be pointed in the SAME direction. You can then say that the third charge is located a distance x to the right of the negative charge.

F1 = kQ1Q/r1^2
F2 = kQ2Q/r2^2

substituting for r1, r2, and dividing by k:
Q1/(L + x)^2 = Q2/x^2

You can solve numerically, getting that x = .91 m or -.11 m. Since the -.11 m location puts the charge between the first two (which we previously said could not happen), only the x = .91 m solution works.

Important Information Alert!
Notice that in 27, 45, and 56, it is stated in the problem that electric field is constant. If you skim through the problem too quickly, you may miss this fact.

27. F = qE = ma so E = 7.10 x 10^-10 N/C. Since it is an electron, the electric field will be directed in the opposite direction of the acceleration, which would be to the south.

45. The weight of the water drop is balanced by the electric force.

mg - qE = 0.

To find m, you must use the volume of a sphere, 4/3*pi*r^3, times the density of water, 1000 kg/m^3.

You can then solve for Q, and then divide this value by the charge on an electron (1.6 x 10^-19 C) to find the number of electrons on the drop.

56. Draw a FBD: T + qE - mg = 0. Thus E = (mg - T)/Q = 1.06 x 10⁷ N/C. Since we assumed in our FBD that E was directed upwards, the negative answer means the field is actually directed downwards.

There Is No Electric Field inside Tom - He's the Conductor!

Electric Field solutions:

p 496 - 498, Q 18, 19, P 33, 34, 35, 38, 43

1981B3 Solutions:
b) T cos(th) - mg = 0. T = mg/cos(th) = 5.65 x 10^-2 N

T sin(th) - qE = 0. E = T sin(th)/Q = 5650 N/C

c) Since the electric field will accelerate the sphere in the horizontal direction, and gravity will accelerate the ball in the vertical direction, the sphere will travel in a straight line directed downwards.

Questions:

18. The field decreases in strength A, B, and then C.

19. If field lines crossed, it would mean that a positive charge could move in two possible directions at the intersection of the lines. This contradicts the definition of the direction of electric field that we came up with, which was THE direction that a positive charge would move at a location.

33.

34.

35. Draw a FBD: Fnet = qE = ma. You know the mass of a proton from the back of the book, as well as the charge, so you can solve for E = 0.10 N/C.


38. The acceleration from the field comes from qE = ma, such that a = 3.24 x 10¹⁵ m/s^2.

We can use our constant acceleration equations because E, and therefore the electric force are constant. v^2 = vo^2 + 2a*delta_x.

v = 8.83 x 10⁶ m/s

b) The ratio of the forces qE/mg = 3.0 x 10^-15.

43. Again, from equilibrium, qE - mg = 0. E = (1.67 x 10^-27 kg)(9.8 m/s^2)/(1.6 x 10^-19C = 1.02 x 10^-7 N/C directed upwards.

Charging Into The Field - Electric field and 2D Electrostatics

Problems 12, 13, 18, 22, 26, 29

12. Remember the trick using your thumbs to find the direction of the force - same distance for all three, and a 60 degree angle for each force. Don't forget to find the direction - 83.8 N directed AWAY from the center of the triangle.

13. Again, this is not actually as tedious as it could be since it is a square and the charges are all the same, so your answer will end up being the same for each charge. You must use the Pythagorean theorem to find the distance of each charge from the charge on the opposite corner of the square (1.41 m).

For each charge, you will end up with something like Fnetx = F1 + F2 sin 45, and Fnety = F3 + F2 cos 45. (The subscripts will be different for each charge.) The final answer should be 6.2 x 10⁵ N directed away from the center of the square.

18. A little mix of Newton's 2nd and electrostatics here - the FBD of the electron should have kQ1Q2/r^2 directed towards the center of the circular orbit. The acceleration is v^2/R.

Q1 = Q2 = e (the fundamental charge), so ke^2/R^2 = mv^2/R.

Thus R = ke^2/mv^2 = 2.2 x 10^-10 m. It is interesting that this is not actually so far off from the actual size of an atom - you have now used basic physics to derive this fact! Not bad, eh?

22. This is a basic application of the definition of electric field: E = F/q. The magnitude comes from rearranging this definition: F = qE = 5.6 x 10^-16 N. Since it is an electron, which has a negative charge, the force will be directed towards the West, opposite the field.

26. This problem is very much like the one we did in class during 7th period - pretend that there is a small test charge at the midpoint of the two charges. Both E1 and E2 are directed to the left (or away from the positive charge) so their magnitudes should be added together. The net field (using the electric field for a point charge E = kQ/R^2) is 3.2 x 10⁸ N/C.

29. We need to use the SAME concept from the other problems in 2D to solve this problem. Unfortunately it's all algebraic, but fear not. The key is to start with a good diagram. I'm going to take mine from the nice folks at the textbook company:


In the diagram, E_- represents the electric field direction from the negative charge, and the other represents the direction for the positive field.

The distance from the origin is x, so the distance from x to each charge is (x^2 + a^2)^(1/2).

The other trick is to find the angle, and THEN find the sine of the angle, but you don't actually have to break that up into two steps. We know from trigonometric ratios that Sin (theta) = opposite side/hypotenuse. The hypotenuse is (x^2 + a^2)^(1/2) and the opposite side of the angle in the right triangle drawn by E_- is a. this means that sin(theta) = a/(x^2 + a^2)^(1/2).

The net electric field in x will be zero, since the charges are the same. In the y-direction, you should get that E = 2*E_- sin(theta).

Substituting our expression for the sine and simplifying, E = 2kQa/(a^2 + x^2)^3/2

But Officer, You Haven't Charged me with Anything! - Electrostatics

P. 496 - 497: Q1, P1, 2, 8, 11, 20

Q1.

Problems:
1. 1.88 x 10¹⁴ electrons

2. F2 = 2.4 N

8. number of excess electrons = 3.8 x 10¹⁴ electrons, 3.4 x 10^-16 kg increase (since you know the increase of electrons, and the mass of one electron is 9.11 x 10^-31 kg)

11. the three net forces on each F1 = 1.2 x 10²N to the left, F2 = 5.3 x 10²N to the right, F3 = 3.9 x 10² N.

20. If one charge is Q1 , the other charge will be Q2 = Q – Q1 . For the force to be repulsive, the two charges must have the same sign. Because the total charge is positive, each charge will be positive. We account for this by considering the force to be positive:

F = kQ1Q2/r2 = kQ1(Q – Q1)/r2;

You know F = 12 N and r = 1.06 meters. This gives you a quadratic equation that can be solved numerically (on the calculator!) for Q1 = 50.0 x 10^-6 C, Q2 = 30 x 10^-6 C