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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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fomenos

Answer:

The force is calculated as 338.66 N

Explanation:

We know that force is given by

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it is given that R=130 m applying values in the above equation we get

R=\frac{v_{o}^{2}sin(2\theta )}{g}\\\\v_{0}=\sqrt{\frac{Rg}{sin(2\theta )}}\\\\\therefore v_{o}=\sqrt{\frac{130\times 9.81}{sin(2\times 50.5_{o}} )}\\\\v_{o}=36.04m/s

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Thus force equals F=338.66N

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3 years ago
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Answer:

Explained below

Explanation:

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2) Pulleys: An example of this in the human body is the knee cap where the direction of an applied force is changed. Thus means as it is in motion, it alters the direction for which the quadriceps tendon pulls on the tibia.

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A golf ball is launched horizontally at a speed of 11 meters per second and a high of 6.4 m above the ground. How long will it t
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The answer is A.

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Answer:

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3 years ago
A 62 kg boy and a 37 kg girl use an elastic rope while engaged in a tug-of-war on a friction-less icy surface. If the accelerati
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Answer:

The magnitude of the acceleration of the boy toward the girl is 1.31\ m/s^2

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To find,

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Solution,

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a_1=-1.31\ m/s^2

|a_1|=1.31\ m/s^2

So, the magnitude of the acceleration of the boy toward the girl is 1.31\ m/s^2

7 0
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