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3 years ago

Calculate the acceleration of a train travelling from rest to 24 m/s in 12 seconds.

Physics

1 answer:

AfilCa [17]3 years ago

3 0

Answer:

2 m/s²

Explanation:

Acceleration is equal to velocity divided by the time taken.

24 / 12

= 2

The units for the acceleration is m/s².

The acceleration of the train traveling from rest to 24 m/s in 12 seconds is 2 m/s².

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The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

$F=G\frac{Mm}{r^2}$

where

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r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

$F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N$

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s$

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