If a 1300-kg car can accelerate from 35 kmh to 65 kmh in 3.8 s, how long will it take to accelerate from 55 kmh to 95 kmh? Assume the power stays the same, and neglect frictional losses.

6.76 s

Explanation:

Power = Energy/time

P = E/t Equation 1

Where P = Power, E = kinetic energy, t = time.

But,

E = 1/2m(Δv)² Equation 2

Where m = mass of the car, Δv = change in velocity.

Substitute equation 2 into equation 1

P = 1/2m(Δv)²/t Equation 3

Given: m = 1300 kg, Δv = 65-35 = 30 km/h = 30(1000/3600) m/s = 8.33 m/s, t = 3.8 s

Substitute into equation 3

P = 1/2(1300)(8.33²)/3.8

P = 11869.15 W.

Assuming the power stays the same,

making t the subject of formula in equation 3

t = 1/2m(Δv)²/P Equation 4

Given: m = 1300 kg, Δv = 95-55 = 40 km/h = 11.11 m/s, P = 11869.15 W

Substitute into equation 4

t = 1/2(1300)(11.11²)/11869.15

t = 6.76 s

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.

that would be because of both gravity and friction. so     c