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Semmy [17]
3 years ago
14

The basic SI unit of length is

Physics
1 answer:
jekas [21]3 years ago
5 0

Answer:

<u>m</u>

Explanation:

<em>Metres</em><em>,</em><em> </em><em>m</em><em>,</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>SI</em><em> </em><em>unit</em><em> </em><em>of</em><em> </em><em>length</em><em> </em>

Hope this helped you-have a good day bro cya)

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You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with
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Answer:

Sam's and Abigail speeds before colliding were a. 12.34 m/s and b. 2.86 m/s, respectively. Their total kinetic energy was diminished by c. 1484.42 J, approximately

Explanation:

By conservation of momentum, we have

\Delta \vec{p} = 0\\\\m\vec{v}_i = m\vec{v}_f\\m_S v_S\hat{i} + m_A v_A\hat{j} = m_S \times 7 \cos{(40)} \hat{i} + m_S \times 7 \sin{(40)} \hat{j} + m_A \times 9.4 \cos{(18)} \hat{i} - m_A \times 9.4 \sin{(18)} \hat{j}.

Writing for each direction at a time,

m_S v_S = 7m_S \cos{(40)} +9.4 m_A \cos{(18)}\\v_S = 7 \cos{(40)} + 9.4 \frac{m_A}{m_S} \cos{(18)} = 7 \cos{(40)} + 9.4\times\frac{57}{73} \cos{(18)} \approx \mathbf{12.34}\\m_A v_A = 7m_S\sin{(40)} - 9.4m_A\sin{(18)}\\v_A = 7\frac{m_S}{m_A}\sin{(40)} - 9.4\sin{(18)} = 7\times\frac{73}{57}\sin{(40)} - 9.4\sin{(18)} \approx \mathbf{2.86}.

Their kinetic energy changed by

K_f - K_i = \left( \frac{1}{2}m_s v_{fs}^2 + \frac{1}{2}m_a v_{fa}^2 \right) - \left( \frac{1}{2}m_s v_{is}^2 + \frac{1}{2}m_a v_{ia}^2 \right) =  \left( \frac{1}{2}\times 73 \times 7^2 + \frac{1}{2}\times 57 \times 9.4^2 \right) - \left( \frac{1}{2}\times 73 \times 12.34^2  + \frac{1}{2}\times 57 \times 2.86^2 \right) \approx \mathbf{-1484.418 J}.

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