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Norma-Jean [14]
3 years ago
11

A 15 kg ball is thrown into the air. It is going at 4m/s when thrown. How much potential energy will it have at the top?

Mathematics
1 answer:
ira [324]3 years ago
3 0

Answer:

P.E = mgh = 120N

Step-by-step explanation:

v² = u² - 2as

At the top, the velocity is = 0

0² = 4² - 2 x 10 x s

4² = 20s

s = 16/20

s = 0.8metres.

It rose only 0.8metres.

Therefore P.E = 15 x 10 x 0.8

:. P.E = 15 * 8

:. P.E = 120N

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Answer:

we conclude that:

3\sqrt{125}+4\sqrt{20}=23\sqrt{5}    

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Step-by-step explanation:

Given the expression

3\sqrt{125}+4\sqrt{20}

Combining the radical expressions

3\sqrt{125}+4\sqrt{20}

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similarly solving

4\sqrt{20}

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Apply radical rule:  \sqrt{ab}=\sqrt{a}\sqrt{b},\:\quad \:a\ge 0,\:b\ge 0

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4\sqrt{20}=8\sqrt{5}

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3\sqrt{125}+4\sqrt{20}=23\sqrt{5}    

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