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Tanzania [10]
2 years ago
14

1. A large ball was let go on a hill and started rolling down with a constant acceleration of 4.2 m/s². What was the velocity of

the ball after 12 s?​
Physics
1 answer:
pochemuha2 years ago
8 0

Answer:

<em>The velocity after 12s is 50.4m/s</em>.

Explanation:

<em>In acceleration formula make velocity the </em><em>subject.</em>

<em> acceleration(a) = velocity(</em>v)÷time(t)

<h3><em> </em><em>velocity</em><em> </em><em>(</em><em>v)</em><em> </em><em>=</em><em> </em><em>acceleration</em><em>(</em><em>a)</em><em>×</em><em>t</em><em>ime</em><em>(</em><em>t)</em></h3>

<em>V </em><em>=</em><em> </em><em>4</em><em>.</em><em>2</em><em>m</em><em>/</em><em>s²</em><em>×</em><em>1</em><em>2</em><em>s</em>

<em>V </em><em>=</em><em> </em><em>5</em><em>0</em><em>.</em><em>4</em><em>m</em><em>/</em><em>s</em>

<em>Therefore</em><em> the</em><em> </em><em>velocity</em><em> </em><em>after</em><em> </em><em>1</em><em>2</em><em>s</em><em> </em><em>is </em><em>5</em><em>0</em><em>.</em><em>4</em><em>m</em><em>/</em><em>s.</em>

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Answer:

a.After 15 second Mr Comer's speed =  1.7 m/s

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Explanation:

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V_{f} =V_{i} + at

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Then Mr.Comer's speed after 15 sec = 1.7ms^-1

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V_{f} ^2-V{i}^2 = 2as

So s = distance covered.

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\frac{V_{f} ^2-V{i}^2}{2a} = 2as

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\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as

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So Mr.Comer will travel a distance of s= 18.75\ m.

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