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charle [14.2K]
2 years ago
5

You are locked inside the train car and want to get it moving to draw attention to your plight. There is effectively no friction

between the axle and the car, and the train is on horizontal tracks. To try and get the car moving with respect to the ground, you run and slam with all your force against the wall at the front. What happens with the car after you slammed against the wall of the car
Physics
1 answer:
AnnZ [28]2 years ago
7 0

Answer:

the car movves briefly as you ran, however, it stops again after you ran in to the wall

Explanation:

Our net total linear momentum was zero at the time both the train and the boy was at rest. As there is no pressure, the train and boy system's linear momentum can be conserved provided no external forces are working on it that might shift its momentum.

If the boy runs inside of the train with a velocity V1 in the forward direction, the train would have a velocity V2 in the reverse direction to V1 to conserve the system's momentum, resulting in Final momentum.

i.e

m \times V_1 + M \times V_2 = 0  --- (1)

here;

m = mass of the boy

M = mass of the train

Thus;

m \times V_1 =- M \times V_2 --- (2)

As the boy crashes into the train's wall, a pair of equal and opposing force F intervene on both the train and the boy. This force F causes the boy traveling with momentum m× V1 to come to a halt; its momentum remains zero. As the train moves with about the same momentum as the boy as seen in (2) and is subjected to the same force F, its momentum will be diminished to zero, and it will come to a halt.

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Answer:

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The equation r (t )=(2t + 4)⋅i + (√ 7 )t⋅ j + 3t ²⋅k the position of a particle in space at time t. Find the angle between the v
velikii [3]

Answer:

\theta = n\pi/2, {\rm where~n~is~an~integer.}

Explanation:

We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

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Similarly,

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\vec{v}(t).\vec{a}(t) = |\vec{v}(t)|.|\vec{a}(t)|.\cos(\theta)\\36t = \sqrt{4 + 7 + 36t^2}.6.\cos(\theta)

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where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

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