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8090 [49]
2 years ago
9

Please help have due very soon?thank you

Physics
1 answer:
Andru [333]2 years ago
8 0
The answer is :78 I think
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Large, above the ground mausoleums were not common so the answer is option B.

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Two ice skaters, Daniel (mass 70.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while a
wel

Answer:

a) v=7.32m/s

b) \alpha =-35º

c) ΔK=-1094.62J

Explanation:

From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together

So, If we analyze the collision we got

p_{1x}=p_{2x}

To simplify the problem, lets name D for Daniel and R for Rebecca

a) p_{D1x}+p_{R1x}=p_{D2x}+p_{R2x}

Since Daniel's initial velocity is 0

m_{R}v_{Rx}=m_{D}v_{D2x}+m_{R}v_{R2x}

v_{D2x}=\frac{m_{R}*v_{R1x}-m_{R}*v_{R2x}}{m_{D}}

v_{D2x}=\frac{(45kg)(14m/s)-(45kg)(8cos(55.1)m/s)}{(70kg)}=6m/s

Now, lets analyze the movement in the vertical direction

p_{1y}=p_{2y}

Since p_{1y}=0

0=m_{D}v_{D2y}+m_{R}v_{R2y}

v_{D2y}=-\frac{m_{R}v_{2Ry}}{m_{D}}=-\frac{(45kg)(8sin(55.1)m/s)}{(70kg)}=-4.21m/s

Now, we can find the magnitude of Daniel's velocity after de collision

v_{D}=\sqrt{(6m/s)^2+(-4.21m/s)^2}=7.32m/s

b) To know whats the direction of Daniel's velocity we need to solve the arctan of the angle

\alpha =tan^{-1}(\frac{v_{y}}{v_{x}})=tan^{-1}(\frac{-4.21}{6})=-35º

c) The change in the total kinetic energy is:

ΔK=K_{2}-K_{1}

ΔK=\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J

That means that the kinetic energy decreases

5 0
2 years ago
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