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course Phy 121
Problem Number 1If 2 gallons of paint are uniformly spread out over the surface of a sphere of radius 7.1 meters, how much paint is applied per square meter?
If the radius doubles, by what factor does the amount per square meter change?
If the radius quadruples, what is the factor?
If the radius is 8 meters, what is the factor?
A = 4 * pi * r^2
A = 633 m^2
2 gallons / 633 m^2 = 0.003 gal/m^2
Factor of 4 if the radius doubles
Factor of 8 if the radius quadruples
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If the radius becomes 4 times as great, r^2 becomes 4^2 = 16 times as great.
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A = 4 * pi * 8^2 = 804 m^2 and 0.002 gal/m^2
Problem Number 2
How many degrees are in each of the following angles:
1 radian = 57 degrees
pi radians = 180 degrees
pi/2 radians = 90 degrees
pi/6 radians = 30 degrees
Problem Number 3
A rigid rod rotating about the center of the circle holds a mass of 0.9399 kg in a vertical circle of radius 1.689 meters, where it moves at a uniform speed of 1.2 m/s.
Find the force exerted by the rod at three points:
The highest point of its path
The lowest point, and
A point whose altitude is identical to that of the center of the circle
a = v^2 / r = (1.2m/s)^2 / 1.689 m = 0.85 m/s^2
Highest point
-Fnet = -m*g + rod force
Rod force = 8.41 N
Lowest Point
Fnet = -m*g + rod force
Rod force = 10.0 N
Center
Fnet = centF = rod force = 9.21 N
Problem Number 4
At a certain instant a disk is rotation at .2 radians/second. 2.1 seconds later it is rotating at .4 radians/second.
If it was accelerated uniformly as a result of a constant torque of 6.599 meter Newtons, then what is its moment of inertia?
tau = 6.599 mN
alpha = angular acceleration = 0.4 - 0.2 / 2.1 = 0.10 rad/s^2
6.599 mN = (0.10) * I
I = 65.99
Problem Number 5
What are the KE and PE changes and the ratio of PE to KE change if an Earth satellite of mass ' mass kg, originally in a circular orbit of radians 7.25 * 10^6 m, increases its orbital radius to 7.43 * 10^6 m?
g1 = (7.25 * 10^6 / 6.38 *10^6) * 9.8 = 7.59 m/s^2
v1 = sqrt(gr) = 7418 m/s
g2 = (7.43 * 10^6 / 6.38 * 10^6) * 9.8 = 7.23 m/s^2
v2 = sqrt(gr) = 7329 m/s
I don't know how to get the energies after this step.
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Mass is given as `mass, so the KE values would be
KE_1 = 1/2 `mass * (7418 m/s)^2
and
KE_2 = 1/2 `mass * (7329 m/s)^2
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Problem Number 6
Obtain an expression for the gravitational field strength of the Earth at distance r from the center by using proportionality. Use your knowledge that the field strength is inversely proportional to distance from Earth's center, the radius 6400 km of Earth and the field strength 9.8 m/s^2 at the surface.
Give the proportionality constant k, then use your result to calculate the field strength at distances 20000, 200000, and 2000000 kilometers from the center of the Earth.
g = k/r^2
k = (6400)^2 * (9.8)
k = 4.01 * 10^8
g = 4.01 * 10^8 / 20000^2 = 1.00 m/s^2
g = 4.01 * 10^8 / 200000^2 = 0.01 m/s^2
g = 4.01 * 10^8 / 2000000^2 = 1.00 * 10^-4
Problem Number 7
A hoop of radius 11.49 meters and mass 2.5 kilograms is attached to a hub centered at its axis of rotation, located at its center and perpendicular to the plane of the loop, by a massless rod. A single mass of 1.399 kilograms is attached to the rod at a point 6.779 meters from the axis of rotation and a torque of 12.49 meter Newtons is exerted on the system.
What are the moment of inertia and the angular acceleration of the system?
I = sigma(mr^2)
alpha = tau / I
I1 = (11.49)*(2.5) = 28.7 kgm^2
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You didn't include units in your calculation. If you had, and had done the unit calculation, you would realize that the units of this calculation come out kg * m rather than kg * m^2.
This would indicate a fault in your formula for the moment of inertia.
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I2 = (1.399)*(6.779) = 9.48 kgm^2
I total = 38.2 kgm^2
alpha = 12.49 mN / 38.2 kgm^2 = 0.33 rad/s^2
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Except for the error regarding moment of inertia, your work on this problem was good.
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Problem Number 8
A tower begins at the surface of the Earth, at a distance of 6400 km from the center, and rises to a position 800 kilometers further from the center.
How much force would an individual of mass 83 kg need to exert against gravity at the beginning and at the end of the climb, in order to climb this tower?
If the average force exerted was equal to the average of the initial and final force, how much energy would be required to climb the tower?
If the individual was reasonably well-conditioned and capable of an average sustained power output of .8 watt/kg for 8 hours per day, how many days would be required to climb the tower?
F = m*a = (83 kg) * (9.8 m.s^2) = 813 N
Radius ratio = 6400 + 800/6400 = 1.13
g = 9.8 m.s^2 / 1.13 = 8.67 m/s^2
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Gravitation is inverse-square in nature, so the required ratio would be 1/1.13^2, not 1/1.13.
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F = (8.67) * (83) = 720 N
aveF = 766 N
W = 766 * 800000 = 6.13 * 10^8 J
avePower = (83) * (0.80) = 66.4 W
E = 66.4 * 8 hours * 3600 s/h = 1.91 * 10^6 J
Time = (6.13 *10^8 J) / (1.91 * 10^6 j=J/d) = 321 days
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You've made a few errors, but overall you appear to be in very good shape for this test.
Check my notes and let me know if you have questions.
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