Batteries & Internal Resistance

From Handout:

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
24. current = .227 A

Circuit Problem Solutions

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

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!