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denis-greek [22]
3 years ago
5

Isaac drops a rubber ball drom height of 2.0m and it bounces to a height of 1.5m. a) What fraction of it's initial energy is los

t during the bounce? b)What is the rubber ball's speed just before and just after it bounces? c)Where did the energy go?
Physics
1 answer:
True [87]3 years ago
7 0

Answer:

a)  ΔE = 25 %

b) v = 8,85 m/s

c) The energy was used against air resistance

Explanation:

In any situation total energy of a body is equal to potential energy plus

kinetic energy, then, just at the moment when Isaac dop the ball the situation is:

Ei = Ep + Ek         where     Ep = m*g*h    and  Ek = 1/2*m*v²

As v = 0  (Isaac drops the ball)

Ei = Ep  =  m*g*h    = 2*m*g

At the end (when the ball bounced to 1,5 m

E₂ = Ep₂  + Ek₂         again at that point  v =0 and

E₂ = 1,5*m*g*

Ei =  E₂ + E(lost)

E(lost) = Ei - E₂

E(lost) = 2*m*g* - 1,5*m*g      and the fraction of energy lost is

E(lost)/Ei

ΔE = (2*m*g* - 1,5*m*g )/ 2*m*g

ΔE = 0,5*m*g / 2*m*g

ΔE = 0,5/2

ΔE = 0,25      or     ΔE = 25 %

b) The speed of the ball is

Potential energy is converted in kinetic energy just when the ball is touching the ground, then

m*g*h = 1/2*m*v²

2*h*g = 1/2 *v²

v² = 4*g*h

v² = 4*2*9,8

v² = 78,4

v = 8,85 m/s

If the impact is an elastic collision, then Ek before and after the impact is the same.

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BigorU [14]

Answer:

QC = 122 KJ

QH = 2.64 x 122 = 322 KJ

Explanation:

TH = 500 Degree C = 500 + 273 = 773 K

TC = 20 degree C = 20 + 273 = 293 K

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Use the formula for the work done in a cycle

Wcycle = QH - QC

200 = QH - QC    ..... (1)

Usse

TH / TC = QH / QC

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QH / QC = 2.64

QH = 2.64 QC     Put it in equation (1)

200 = 2.64 QC - QC

QC = 122 KJ

So, QH = 2.64 x 122 = 322 KJ

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2 years ago
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