You drop a ball from a height of 1.7 m, and it bounces back to a height of 1.2 m. Part A) What fraction of its initial energy is lost during the bounce? Express your answer using two significant figures.
Part B) What is the ball's speed just before the bounce? Express your answer to two significant figures and include the appropriate units.
Part C) What is the ball's speed just after the bounce? Express your answer to two significant figures and include the appropriate units.
Part D) Where did the energy go?
1. The energy "lost" was changed primarily into chemical energy.
2. The energy "lost" was changed primarily into acoustic energy.
3. The energy "lost" was changed primarily into thermal energy due to friction.
4. The energy "lost" was changed primarily into heat energy.

A)       ΔEm = 0.29,  B)   v₁ = 5.8 m/s, c)   v₂=  4.9 m / s    D) the correct answer from 4

Explanation:

For this exercise we will use conservation of energy, taking care of how to choose our system

A) For this case we take two instants

starting point. When the ball goes out

        Em₀ = U = m g y₁

Final point. When the ball reaches its maximum height

          = U = m g y₂

In this case we see that there is a loss of mechanical energy at the moment of rebound, therefore the fraction of energy lost is

          ΔEm = Em_{f} / Em₀

          ΔEm = mg y₂ / mg y₁

          ΔEm = y₂ / y₁

          ΔEm = 1.2 / 1.7

the lost part of energy  

          ΔEm = 1 -0.706

          ΔEm = 0.29

B) the velocity just before the bounce

starting point. When the ball is released

          Em₀ = U = m g y₁

final punot. Just wide of the bounce

           Em_{f} = K = ½ m v₁²

As it has not yet rebounded, it has no energy loss, therefore the mechanical energy is conserved

            Em₀ = Em_{f}

            m g y₁ = ½ m v₁²

             v₁ = √ 2 g y₁

let's calculate

            v₁ =√ (2 9.8 1.7)

             v₁ = 5.77 m / s

            v₁ = 5.8 m/s

C) the velocity just after the bounce

   starting point, after bounce

               Em₀ = K = ½ m v₂²

   final point. Maximum height after bounce

               Em_{f} = U = m g y₂

as it already bounced, the energy is conserved in this interval

               Em₀ = Em_{f}

               ½ m v₂² = m g y₂

               v₂ = √ (2 g y₂)

               v₂ = √ (2 9.8 1.2)

               v₂ = 4.85 m / s

               v₂=  4.9 m / s

D) during the time that the bounce lasts, there is a strong change in energy, part of it is transformed into thermal energy, due to several processes: friction, change in the potential energy of the molecules of the ball, change in the internal energy of the balls. molecules.

Therefore we cannot specify a single process, consequently the correct answer from 4

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.