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torisob [31]
3 years ago
11

A space vehicle is traveling at 2980 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The

relative speed between the motor and the command module is then 94 km/h. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth just after the separation
Physics
1 answer:
Strike441 [17]3 years ago
8 0

Answer:

3054.4 km/h

Explanation:

Using the conservation of momentum

momentum before separation = 5M × 2980 Km/h where M represent the mass of the module while 4 M represent the mass of the motor

initial momentum = 14900 M km/h

let v be the new speed of the motor so that the

new momentum = 4Mv and the new momentum of the module  = M ( v + 94 km/h )

total momentum = 4Mv + Mv + 93 M = 5 Mv + 93M

initial momentum = final momentum

14900 M km/h = 5 Mv + 93M

14900 km/h = 5v + 93

14900 - 93 = 5v

v = 2961.4 km/h

the speed of the module = 2961.4 + 93 = 3054.4 km/h

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ArbitrLikvidat [17]

Answer:

The value is v_y  =  -48.61 \ m/s

Explanation:

From the question we are told that

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Generally the time taken by the hot magma in air before landing is mathematically represented as

       t = \frac{d}{u_x}

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=>    t = 4.96 \  s

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Then the final velocity of the magma when it hits the ground is mathematically represented s

       - v_y  =  u_y + gt

Here the negative sign mean that the direction of the velocity is towards the negative y -axis

So  

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Explanation:

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Answer:

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Hope this is correct and helpful

HAVE A GOOD DAY!

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