LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust. The roc

Question

LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
The rocket engines switch off at 2.5 km and the rocket slows down as it moves up until it stops momentarily and then falls to the ground.
i Calculate the rocket’s energy when the rocket engines were switched off.
ii Calculate the speed of the rocket when the rocket engines were switched off.

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Kim Chi 4 years 2021-08-14T16:51:09+00:00 1 Answers 27 views 0

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  1. ngocdiep
    0
    2021-08-14T16:52:41+00:00
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    Answer:

    i) E = 269 [MJ]    ii)v = 116 [m/s]

    Explanation:

    This is a problem that encompasses the work and principle of energy conservation.

    In this way, we establish the equation for the principle of conservation and energy.

    i)

    E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

    W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

    At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

    Er = 269*10^6[J]

    ii ) With the energy calculated at the previous point, we can calculate the speed developed.

    E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

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