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kati45 [8]
2 years ago
13

ball is dropped from a height of 1.60 m and rebounds to a height of 1.20 m. Approximately how many rebounds will the ball make b

efore losing 90% of its energy?
Physics
1 answer:
jasenka [17]2 years ago
3 0

Answer:

the ball rebounds 8 times before losing 90% of its energy

Explanation:

given information:

height, h₁ = 1.6 m

rebound height, h₂ = 1.2 m

energy loss = 90% = 0.9

first we can calculate the loss energy after the rebound

energy loss = 1 - [(height of rebound/height of the ball)^n

where

n = number of the bounce

energy loss = 1 - (h₂/h₁)^n

0.9 = 1 - (1.2/1.6)^n

(0.75)^n = 1 - 0.9

n log 0.75 = log 0.1

n = log 0.1/log 0.75

  = 8 bounces

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I made the drawing in the attached file.

I included two figures.

The upper figure shows the effect of:

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In the lower figure you have the resultant vector: C = 1.5A - 3B.

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7 0
2 years ago
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Determine the number of unpaired electrons in the octahedral coordination complex [fex6]3–, where x = any halide.
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3 0
2 years ago
3) A defense football player on one team tackles the other team’s quarterback, who is running down the field. The quarterback is
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2 years ago
If you find an igneous rock which has 450 radioactive isotopes and 3,150 stable daughter isotopes, how many half-lifes of this i
slavikrds [6]

Answer:

3t_{1/2}  

Explanation:

To find the half-lifes of the isotope we need to use the following equation:

N_{t} = N_{0}2^{-\frac{t}{t_{1/2}}}     (1)

<em>where Nt: is the amount of the isotope that has not yet decayed after a time t, N₀: is the initial amount of the isotope, t: is the time and </em>t_{1/2}<em>: is the half-lifes.</em>

By solving equation (1) for t we have:

\frac{t}{t_{1/2}} = - \frac{Ln(Nt/N_{0})}{Ln(2)}

<u>Having that:</u>

Nt = 450

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The half-lifes of the isotope is:

t = - \frac{Ln(450/3600)}{Ln(2)} \cdot t_{1/2} = 3t_{1/2}

Therefore, 3 half-lives of the isotope passed since the rock was formed.

I hope it helps you!

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