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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:

1.38*10^18 kg

Explanation:

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F=G*\frac{m_a*m_p}{r^2}

where:

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ma= mass of the astronaut

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F=m_a.a\\(v_f )^2=(v_o)^2+2.a.\Delta y\\\\a=\frac{(v_f)^2-(v_o)^2}{2.\Delta y}\\\\a=\frac{(0)^2-(4.29m/s)^2}{2.0.64m}=14.38m/s^2\\\\F=m_a*14.38m/s^2

so:

m_a*14.38m/s^2=(6.674*10^{-11}N.(m/kg)^2)*\frac{m_a.m_p}{(2.530*10^3m)^2}\\m_p=\frac{14.38m/s^2(2.530*10^3m)^2}{(6.674*10^{-11}N.(m/kg)^2)}\\\\m_p=1.38*10^{18}kg

7 0
3 years ago
8. At what position does the mass have the greatest acceleration?
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Answer:

Option (e)

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At position +A, mass will have the minimum kinetic energy, zero velocity and maximum acceleration.

Therefore, Option (e) will be the answer.

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The heat energy transferred by the iron nail is 4680 J

Explanation:

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m is the mass of the substance

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\Delta T is its change in temperature

For the iron nail in this problem, we have:

m = 16 g

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So, the amount of heat energy given off by the nail is

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Learn more about specific heat capacity:

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