A runaway railroad car, with mass 4.50 × 105 kg, coasts across a level track at 3.80 m/s when it collides with a spring-loaded bumper at the end of the track. If the spring constant of the bumper is 3.80 × 106 N/m, what is the maximum compression of the spring during the collision? (Assume the collision is elastic.)

1.31 m

Explanation:

applying the law of conservation of energy,

The kinetic energy of the car = energy of the spring loaded bumper

1/2mv² = 1/2ke²

make e the subject of the equation

e = √(mv²/k) Equation 1

Where m = mass of the car, v = velocity of the car, k = spring constant of the bumper, e = maximum compression of the spring during collision

Given: m = 4.5×10⁵ kg, v = 3.8 m/s, k = 3.8×10⁶ N/m

Substitute into equation 1

e = √(4.5×10⁵×3.8²/3.8×10⁶)

e = √1.71

e = 1.31 m

Hence the maximum compression of the spring during collision = 1.31 m

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?

i guess it’s a and b