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enyata [817]
2 years ago
8

A movie stunt double is supposed to run across the top of a train (in the opposite direction that the train is moving) and just

barely jump off before reaching a tunnel, but after running the entire length of the train (starting from the front). If the train is moving at 50 km/hr, is 3 km long and the tunnel is 40 km away from the end (where the stunt double is going to jump from), how fast (in km/hr) will the stunt double need to run
Physics
1 answer:
Rus_ich [418]2 years ago
3 0

Answer:

3.49km/hr

Explanation:

This train is moving at 50km/hr

Distance = length of train + distance of tunnel

= 3 km + 40 km

= 43 km

Time taken by train to get to tunnel

Time = distance / velocity

Time = 43 km/50km perhour

= 0.86

How fast will stunt double have to run?

Velocity = 3km/0.86

= 3.49 km/hr

The stunt double will have to run as fast as 3.49 kilometer per hour

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m_1v_1=m_2v_2

Therefore,

Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)

=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]

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B)  the speed of the sphere with mass 107.0 kg is

v_2=\frac{m_1v_1}{m_2}

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v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s

D) the distance of the centre is proportional to the acceleration

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When the sphere make contact with eachother

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x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r

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The point of contact of the sphere is

32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m

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