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.