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14.10.2022 05:26

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Suppose a rock on earth is dropped in a vacuum and has initial speed of 0 m/s. what

Suppose a rock on earth is dropped in a vacuum and has initial speed of 0 m/s. what is the rock's speed after 2 seconds, in m/s, if it encounters no air resistance

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CrusaderLord

4,8(92 marks)

What is the rock's speed after 3 seconds, in m/s, if it encounters no air resistance? ... Well, we can use the falling objects equation, which states that. d=12>2. where g is the gravitational acceleration, t is the time in seconds. So, plugging in values into the equation, we get. d=12⋅9.81 m/s2⋅9 s2.

Explanation:

Crxymia

4,9(60 marks)

For the bullet, $m = 19 \text{ g} \cdot \dfrac{1\text{ kg}}{1000 \text{ g}} = 0.019\text{ kg}$ the net force on the bullet: $\sigma \vec{f} = \vec{f}_f = 50 000 \text{ n}$ an fbd would look like a dot with a horizontal "friction" force protruding from one end.by newton's 2nd law: $\sigma \vec{f} = m\vec{a}$ we can solve for acceleration since we know $\sigma \vec{f} = 50000\text{ n}$ and $m = 0.019\text{ kg}$ . $\begin{aligned} \sigma \vec{f} & = m\vec{a}\\ \vec{a} & = \dfrac{ \sigma \vec{f} }{m} \\ \vec{a} & = \dfrac{50000\text{ n}}{0.019\text{ kg}} \\ \vec{a} & = 2631578.95 \text{ m/s}^2 \end{aligned}$ we know that $v_f = 0\text{ m/s}$ (final velocity is bullet not moving), $v_i = 334\text{ m/s}$ . we can use the following kinematic equation to get $d$ , or the displacement. $\begin{aligned} v_f^2 & = v_i^2 + 2ad \\ v_f^2 - v_i^2 & = 2ad \\ d & = \dfrac{v_f^2 - v_i^2}{2a} \\ d & = \dfrac{0 - (334\text{ m/s})^2}{2\cdot 2631578.95 \text{ m/s}^2} \\ d & = -0.02119563997 \text { m} \\ d & = -0.02 \text { m} & & \text{1 significant figure} \end{aligned}$ so it penetrates $0.02$ m.

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