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A 1.47-newton baseball is dropped from a height of 10.0 meters and falls through the air to the ground. The kinetic energy of the ball is 12.0 joules the instant before the ball strikes the ground. The maximum amount of mechanical energy converted to internal energy during the fall is_________.

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jolis1796

Answer:

ΔEinternal = - 2.7 J

Explanation:

Given

W = m*g = 1.47 N

hinitial = 10.0 m

Kfinal = 12.0 J

We can apply

ΔE = ΔEinternal

then

ΔE = (Kfinal + Ufinal) - (Kinitial + Uinitial)

⇒   ΔE = (Kfinal + 0 J) - (0 J + Uinitial) = Kfinal - Uinitial

⇒   ΔE = Kfinal - (W*hfinal) = 12 J  - (1.47 N*10.0 m)

⇒   ΔE = 12 J - 14.7 J = - 2.7 J

then

ΔEinternal = - 2.7 J

The maximum amount of mechanical energy converted to internal energy during the fall is - 2.7 J.

Given :

Baseball Weight = 1.47 N

Height = 10 m

Kinetic Energy (KE) = 12 J

Solution :

We know that the change in total energy is,

[tex]\rm \Delta E = (KE_f-PE_f) - (KE_i-PE_i)[/tex]

[tex]\rm \Delta E = (KE_f-0) - (0-PE_i)[/tex]

[tex]\rm \Delta E = 12-(1.47\times10)[/tex]

[tex]\rm \Delta E= -2.7\;J[/tex]

The maximum amount of mechanical energy converted to internal energy during the fall is - 2.7 J.

For more information, refer the link given below

https://brainly.com/question/18461965