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denis-greek [22]
3 years ago
7

How much potential energy does a 40-N medicine ball gain when it is lifted 5 m?

Physics
1 answer:
777dan777 [17]3 years ago
6 0
We know, Potential Energy = Force * Height
Here, F = 40 N
h = 5 m

Substitute their values, 
U = 40 * 5
U = 200 J

In short, Your Answer would be Option A

Hope this helps!
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Answer

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\large{\maltese{\textsf{\underline{To find :-}}}}

The X and Y coordinates of the rocket relative of firing

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\large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^{-1} \\ \sf angle ({\theta}_{1}) = 55.0{\degree} \\ \sf time (t) = 12s

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\Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}}

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<u>The</u><u> </u><u>horizontal</u><u> </u><u>range</u><u> </u><u>of</u><u> </u><u>projectile</u><u> </u><u>at</u><u> </u><u>x</u><u>.</u><u> </u>

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\sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t}

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\large\textsf{\underline{Now substituting the required values}}

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\sf x = 300 \times \cos 55{\degree} \times 12 \\ \\ \sf x = 100 \times 0.5756 \times 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}}

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The vertical position of projectile at y.

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\sf \large  y = v_{yi} \times t -  (\frac{1}{2}  \times g  \times {t}^{2}) \\  \\   \sf  \large y = v_i \times  \cos \theta  \times t -  \frac{1}{2} g {t}^{2}

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\textsf{ \large {\underline{Now substituting the required values}}  }

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\sf y = 100 \times  \cos55{ \degree} \times 12 -  \frac{1}{2}   \times 9.80 \times  {12}^{2} \\  \\  \sf  y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\  \\  \sf y = 983.04 - 705.6 \\  \\  \underline{ \boxed{ \bold{y = 277.44m}}}

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I might be wrong, but. The words seem to fit. 

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