1. Consider the ball example in the introduction when a ball is dropped from 3 meters. After the ball bounces, it raises to a height of 2 meters. The mass of the ball is 0.5 kg a. Calculate the speed of the ball right before the bounce. b. How much energy was converted into heat after the ball bounced off the ground

(a) 7.67 m/s.

(b)  4.9 J

Explanation:

(a) From the law of conservation of energy,

P.E = K.E

mgh = 1/2(mv²)

therefore,

v = √(2gh) Equation 1

Where v = speed of the ball before bounce, g = acceleration due to gravity, h = height from which the ball was dropped.

Given: h = 3 m, g = 9.8 m/s²

Substitute into equation 1

v = √(2×9.8×3)

v = √(58.8)

v = 7.67 m/s.

(b) Energy of the ball before the bounce = mgh = 0.5×9.8×3 = 14.7 J

Energy of the ball after the bounce = mgh' = 0.5(9.8)(2) = 9.8 J

Amount of energy converted to heat = 14.7-9.8 = 4.9 J

when a device is used in a circuit in which the voltage is 81 v, the current flowing through the device is 3 a. what is the resistance of the device?

a. 27 ω

b. 40.5 a

c. 243 ω

d. 243 a

the correct answer is a. 27