Pls help me with this

Physics

1 answer:

Ilia_Sergeevich [38]3 years ago

6 0

U is correct on the first one

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$\\ \mathrm{Given:} \\\mathrm{Mass \ of \ first \ car}$\left(m_{1}\right)=1383 \mathrm{~kg}$ \\ Velocity $\left(\overrightarrow{V_{1}}\right)=-11.2\ {\math} \mathrm{m} / \mathrm{s}$\\ Mass of second car -$\left(m_{2}\right)=1732 \mathrm{~kg}$\\ Velocity $\left(\vec{v}_{2}\right)=31.3 {\math} \mathrm{m} / \mathrm{s}$$

$m_{1} \vec{v}_{1}+m_{2} \vec{v}_{2}=\left(m_{1}+m_{2}\right) \vec{v} \\1383(-11.2 {\math})+1732(31.3 {\math}) \\=(1383+1732) \vec{v} \\-15489.6 {\math}+54211.6 {\math}=3115 \vec{v} \\ \vec{v}=(17.4 {\math}-4.972 {\math}) \ \mathrm{m/s}$

$\text { So magnitude } \vec{|v|} =\sqrt{(17.4)^{2}+(-4.972)^{2}} \\ =\sqrt{327.605} \\ =18.099 \mathrm \ {m/s} \\$

$\text { Direction } \\\theta=\tan ^{-1}\left(\frac{-4.972}{17.4}\right) \\\theta=-15.945^{\circ}\end{gathered}$

Therefore, both cars move with a velocity of 18.099 m/s in the direction of 15.945° downward from the x-axis (east)

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3 years ago

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