Upon its return voyage from a space mission, the spacecraft has a velocity of 25000 km/h at point A, which is 7130 km from the c
enter of the earth. Determine the velocity of the spacecraft when it reaches point B, which is 6625 km from the center of the earth. The trajectory between these two points is outside the effect of the earth’s atmosphere. The radius of the earth is 6371 km, and g = 9.825 m/s2 on the surface of the earth.
1 answer:
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Answer:
-6112.26 J
Explanation:
The initial kinetic energy, is given by
} where m is the mass of a body and
is the initial velocity
The final kinetic energy, is given by
where
is the final velocity
Change in kinetic energy, is given by
Since the skater finally comes to rest, the final velocity is zero. Substituting 0 for and 12.6 m/s for
and 77 Kg for m we obtain
From work energy theorem, work done by a force is equal to the change in kinetic energy hence for this case work done equals <u>-6112.26 J</u>
Answer:
Rms speed of the particle will be
Explanation:
We have given mass of the air particle
Gas constant R = 8.314 J/mol-K
Temperature is given T =
We have to find the root mean square speed of the particle
Which is given by
So rms speed of the particle will be