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course Phy201
Problem 1How much paint is applied per square meter if 10 gallons of paint are uniformly spread out over the surface of a sphere of radius 8.1 meters?
If the paint is applied over a sphere of double the radius, by what factor does the amount per square meter change?
If the paint is applied over a sphere of four times the radius, by what factor does the amount per square meter change?
If the paint is applied over a sphere of radius 19 meters, what is the factor by which the amount per square meter changes?
Solution
radius r is 4 `pi r ^ 2 area of first sphere = 120.763 m ^ 2
paint = 2 gallons / 120.763 m ^ 2 = .0165 gal/m ^ 2
density on second sphere = .25 ( .0165 gal/m ^ 2) = .00414 gallons/m ^ 2
density on third sphere = 1/16 ( .0165) gal/m^2 = .00103 gal/m^2
density on final sphere = .0165/( 6/ 3.1) ^ 2 gal/m ^ 2 = .00442 gal/m ^ 2
density is .266 times the original amount per square meter
Problem 2
When covered with a coat of paint having density 6 gallons / m^2, a sphere of radius 2.2 meters requires a certain total amount of paint.
If a sphere of radius 11.45 meters is covered with an equal amount of paint, what will be its surface density?
Solution
6 gal/m ^ 2 = k/( 2.2 meters) ^ 2
k = 29.04 gallons
y = 29.04 gal / ( 11.45 meters) ^ 2 =
.22 gal/meter ^ 2
Problem 3
The gravitational field strength at the surface of the Earth is g = 9.8 m/s^2; at the surface of the moon g is 1.5 m/s^2.
Suppose your mass is 64 kg.
What is your weight on the surface of the Earth and on the Moon?
What is the strength of the gravitational field on the surface of a planet where you weigh 2000 Newtons?
Solution
9.8 m/s ^ 2)( 64 kg) = 627.2 Newtons
1.5 m/s ^ 2)( 64 kg) = 96 Newtons
2000 Newtons/ 64 kilograms = 31.25 meters/second ^ 2
Problem 4
Using proportionality, find:
The field strength at twice the radius of the Earth from its center.
The strength at 4 times the radius of the Earth from its center.
The strength at a distance of 22100 kilometers from its center.
The distance from Earth's center where the gravitational field of the Earth is half its value at the surface.
You may use the information that the acceleration of gravity at the surface of the Earth is 9.8 m/s^2 and the the radius of the Earth, which is approximately 6400 km.
Solution
1/4 * (9.8 m/s ^ 2) = 2.45 m/s ^ 2
1/16(9.8 m/s ^ 2) = .6125 m/s^2
ratio of radii = (6400 / 22100 )
field strength at 22100 km = 9.8 m/s ^ 2 * (6400 / 22100 ) ^ 2 = .82 m/s ^ 2
9.8 m/s ^ 2 * (6400 km / R) ^ 2 = 4.9 m/s ^ 2
Problem 5
Calculate the field strength of the Earth's gravitational field at distances 20000, 200000 and 2000000 kilometers from the center of the Earth, using only proportionality and the following facts:.
Field strength is inversely proportional to distance from Earth's center.
The radius of Earth is 6400 km.
Field strength 9.8 m/s^2 at the Earth's surface.
Solution
g = [ 4.01408E+08 km m / s ^ 2] / r ^ 2
g = 4.01408E+08 kg m /s ^ 2 / ( 20000 km) ^ 2 = 1.003 m/s ^ 2
Problem 6
Give the strength of the gravitational attraction felt by a human being of mass 83 kg to a spherical object with radius 2 km and uniform density 3.7 times that of water (water's density is 1000 kg/m ^ 3), assuming that the person's entire mass is located at the surface of the sphere. You may use G = 6.67 * 10^-11 N m^2 / kg^2.
Give the attraction if the sphere was compressed to a radius of 200 meters, and if it was compressed to radius 20 meters.
To what radius would the sphere have to be compressed in order to exert a force equal to the weight of this individual on the surface of the Earth?
Solution
F = G m1 m2 / r ^ 2
G = 6.67 * 10^-11 N m ^ 2/kg ^ 2 to obtain force
F = .171 Newtons.
.8 m/s ^ 2 ( 83 kg) = 813.4 Newtons
r = `sqrt[G m1 m2 / F
r = 29.04963 meters
Problem 7
A person of mass 80 kg begins climbing a very high tower. The 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 does the individual exert against gravity at the beginning and at the end of the climb?
If the average force exerted was equal to the average of the initial and final force, how much energy would be required?
At an average power output of .95 watt/kg for 8 hours per day, how many days would be required?
Solution
F = 80 kg (9.8 m/s ^ 2) = 784 Newtons
radius ratio: (6400 + 800) km / 6400 km
grav field at altitude: g = (9.8 m/s ^ 2) / [(6400+ 800)/6400] ^ 2
gravitational force (weight) at 800 km altitude = 80 kg (9.8 m/s ^ 2) / [(6400+ 800)/6400] ^ 2 = 619.45 Newtons
average of forces = 701.725 Newtons
energy required = ( 701.725 Newtons)( 800000 meters) = 5.6138E+08 Joules
energy in 8 hour working day = ( 76 Joules/second) (8 hours) (3600 seconds/hour) = 2188800 Joules
time required = 5.6138E+08 Joules/( 2188800 Joules/day) = 256.4 days"
&#Good responses. Let me know if you have questions. &#
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