10th. HS - Review - 4th. Unit

The earth and the sun are 1.5 E11 meters apart. Use Newton's Law of Universal Gravitation to calculate the force of gravity between the earth (mass = 6.0 E24 kg) and the sun (mass = 2.0 E30 kg).

• Fg = 2.6 E22 N

• Fg = 1.4 E22 N

• Fg = 3.6 E22 N

• Fg = 3.6 E19 N

Jenny loves puffed rice cereal. As she is pouring the cereal, two little pieces of cereal each with equal charges of E−18 coulombs fall into the bowl 0.01.8525 meters apart. Use Coulomb's Law to find the electrostatic force between them.

• Fe = 2.0 E−23 N

• Fe = 4.9 E−23 N

• Fe = 4.9 E−25 N

• Fe = 6.3 E−23 N

A 100. N sled with steel runners is on a horizontal steel track.

The coefficient of static friction between steel on steel is 0.45. What is the minimum horizontal force required to start the sled to slide along the horizontal track?

• Ff = 55 N

• Ff = 40 N

• Ff = 50 N

• Ff = 45 N

A 100. N sled with steel runners is on a horizontal steel track.

The coefficient of kinetic friction between steel on steel is 0.25. What is the magnitude of the force needed to keep it moving at a steady pace?

• Ff = 25 N

• Ff = 30 N

• Ff = 35 N

• Ff = 55 N

•The mass of the space shuttle is about 5.06 E5 kilograms. What would be the weight of the space shuttle as it sits on the launch pad? Be sure to show your work.

• W = 5.0 E8 N

• W = 6.0 E6 N

• W = 5.0 E6 N

• W = 4.0 E7 N

¡Assume that the shuttle is orbiting 3.22 E5 meters above the surface of the Earth. Use Newton's Law of Universal Gravitation to calculate the weight of the shuttle as it orbits at this height. Hint: Remember that the "d" in the equation is the distance between the center of the shuttle and the center of the Earth. (Earth’s mass = 6.0 E24 kg) ( Earth’s radius = 6.377  E6 m) (Space shuttle = 5.06 E3 kg)

• F = 5.5 E6 N

• F = 4.5 E6 N

• F = 3.5 E5 N

• F = 2.7 E4 N