Parallel and Combination Circuit HW

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.

Ohm's Law and Series Circuits HW

From Handout:

2. 1200 Ohms
3. 10 V, 20 Ohms
4.
b)60 Ohms,
c)0.2 A
d) 15 Ohm - 3 V, 21 Ohm - 4.2 V, 24 Ohm - 4.8 V
e) .6 W, .84 W, .96 W
f) .6W + .84 W _ ,96 W = 2.4 W OR (.2 A * 12 V) = 2.4 W
g) thickest filament - smallest resistance, since it has a larger cross sectional area. This must be the 15 Ohm resistor.
h) The brightest bulb uses the most power - this is the 24 Ohm bulb.

From book:
p. 551:
5. 35.5 Ohms
7. 2.1 x 10²¹ electrons
9. 24 Ohms, 360 W
26. Since P = V²/R, and V is the same for both, the resistance must be less for the higher power. Assuming a wall voltage of 120 V, the resistances are 24 Ohms and 12 Ohms for the 600 W and 1200 W settings.
31.
a) 8.57 Ohms, 1.05 W
b) 4 times the power. The bulb would probably not last for long!

p. 581
10. 27 Ohms

Electric Current and Resistance HW

Solutions to Friday Worksheet:

1. 600 C, 600 W.
2. 4 hours, 10 hours
3. 133.3 A
4. A flow of current through one hand and out the other passes through the heart. The current running through the hand and out the elbow would be painful, but not necessarily fatal.
5. Power of the hairdryer is 12,0000J/45 seconds = 266.7 W, voltage = P/I = 88.9 Volts
6. .034 .34 Ohms
7. B, resistance of a cylindrical wire = p*L/A with p = resistivity, L = length of the wire, and A = cross sectional area.
8. 24 Ohms
9. 60/7 Ohms

34. Three lightbulbs.
35. efficiency = 373 W/528 W (1 hp = 746 W. What is the actual power of the motor? What is the electrical power provided to the motor? How can you use efficiency from thermodynamics to relate these two?)

Midyear Review Answers

I'll be updating this post with answers throughout the night and day, so check them when you can.
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) 25 m/s
(c) 50 m/s^2

6.
(a) 40 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
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.
(a) T1 = 60 N, T2 = 100 N, minimum value of static friction coefficient = .27
(b) T1 = 59 N, T2 = 98 N, a = .17 m/s^2
5.
(a) 22.5 m
(b) 2.5 m
(c) Snoopy is going fast enough in the horizontal direction that the normal force of the plane acting on him holds him in a circular path, rather than a parabolic one.
6. 24 m/s


Midyear Review #4 - Work and Energy
1. 23.8 J
2. 0.8 m
3. 27 meters
4. (Note that the height of the building should be 80 meters, not 30 as written on the sheet.)
(a) 400 J
(b) 5 N
5.
(a) 1400 J
(b) 60 W
6.
(a) 0 J
(b) 8 J
(c) 64 J
7.
(a) 230 J
(b) 10.0 m/s



Midyear Review #5 - Gravity, Momentum, Torque
1.
(a) Mass is the same, 70 kg
(b) 700 N (16) = 11,200 N
(c) 160 m/s^2
2. (GM/r)^(1/2)
3. (1/5)^(1/2)
4. 70 kg
5.
(a) (3/10)v0.
(b) -21/20mv^2
6.
(a) 1.183 m/s
(b) 782 m/s
7.
(a)(.4 kg)(2 m/s) + 0 = (.4 kg)(1.2 m/s) cos 30 + (.4 kg)Vbx
0 = (.4 kg)(1.2 m/s) sin30 + (.4 kg)Vby
(b) Vbx = 0.96 m/s, Vby = -.6 m/s, so Vb = (.96^2 + .6^2)^(1/2) = 1.13 m/s
tan(x) = .6/.96 so x = 32 degrees.
Final velocity of ball b is 1.13 m/s at 32 below the horizontal.
(Ignore the part that says the collision is elastic - if you calculate change in KE, it is NOT zero!)
8.
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

Fluid mechanics answers

Here are the solutions to the Bernoulli problems from Friday:

3) 2.05 x 10^5 Pa

4) a. 64.5 Pa
b. 258 N

Also , here is the URL for the National Youth Science Camp which is offering the summer program I mentioned in class. Applications can be found at http://www.emsc.nysed.gov/ciai/mst/science/natyouth.html

See you all tomorrow!

See the ISS/Shuttle in the morning!

Hi everyone,

As we discussed last week in class, the International Space Station and Space Shuttle are docked together in orbit around the Earth.

At 5:52 tomorrow morning, the International Space Station and Space Shuttle will be visible in the South-western sky until around 5:57. Look for a brightly shining object moving quickly across the sky to the Northeast. At its maximum height, the pair will be at 57 degrees above the horizon.

Check it out!

Work/Energy Solutions

Hello everyone,

This week is going to require that we apply MOST of what we have learned thus far about Newton's Laws and Work/Energy. Working hard is required to really understand how it all comes together.

First, the solutions to the handout problems from class:

1.
a) Determine the magnitude and direction of the force that the spring exerts on the block. [45 N directed to the right.]
b) If the force holding the block against the spring is removed, calculate the work done by the spring on the block. [W_spring = 1/2kx² = 1/2(150 N/m)(0.3m)² = 6.75 J]
c) Calculate the speed of the block when it loses contact with the spring. [Using Work-Energy principle, W_net = KEf - KEi so v_f = 2.598 m/s.
d) If the compression fo the spring is doubled, how would your answer to (c) change? [final speed would double. ]

2.
a) Calculate the displacement of the spring from its equilibrium position. [.049 m]
b) Calculate the work done by gravity as the spring stretches.
c) Calculate the work done by the spring as the spring stretches. [.1801 J]
d) What is the net work done on the mass? [Wnet = 0 since the change in kinetic energy is zero!]
e) How much work was done in lowering the mass to the equilibrium position?

3.
a) Wnet = -24 J
b) Wspring = -1/2kx²
c) Wfric = -mu*mg*deltaX
d) -1/2kx² -mu*mg*deltaX = -24 J so deltaX = .325 m after solving numerically.

Homework Solutions/hints:
Questions:

5. Friction CAN cause an object to accelerate as in the textbook example of ripping out a tablecloth from under dishes. What does this mean about the change in KE? Wnet?

7. Using Uspring = 1/2kx²: (a) Spring 1 - displacement x = F/k. (b) Spring 2

8. KE = 1/2mv², where v is the speed and m is the mass for the object. Neither mass nor v² can be negative, so the answer is no.

9. The amount of work added the same in moving from B to C as compared with the work from A to B. The net work at point C, therefore, is twice the net work at point B. Using the Work-Energy principle, this means that the final speed is the square root of two times vb at point C.

Problems:
11. (a) 1.1*Mg (b) 1.1*Mg*h

14. Using area of a trapezoid: W = .5*(88N/m*.038m+88N/m*.058 m)*(.058m - .038m) = 0.08448 J

29. 0.337 meters

30. mgh = (6 kg)(10 m/s²)(1.2 m) = 72 J

33. a) mgh = (55 kg)(10 m/s²)(3100 m - 1600 m) = 825,000 J
b) minimum work required for the hiker to reach this new height IS the change in PE, or 825,000 J.
c) If the hiker also has a change in KE as compared with the KE at 1600 m, then the work could be more by the work-energy principle.