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serg [7]
2 years ago
9

A 3000-kg spaceship is moving away from a space station at a constant speed of 3 m/s. The astronaut in the spaceship decides to

return to the space station by switching on engines that expel fuel so that the sum of the forces exerted on the spaceship by the expelled fuel points toward the space station. What is the magnitude of the minimum force needed to bring the spaceship back to the space station
Physics
1 answer:
victus00 [196]2 years ago
4 0

Answer:

P = m (v2 - v1)     change in momentum

P = - m v1  if v2 is to be zero

F = P / t     change in momentum equals applied force

F = m a = - m v1 / t

a = -v1 / t     or a * t = -v1`

Any combination of a and t that is greater than v1 will bring the space station to a halt

If a is small then t must be large

Any applied force will eventually bring the ship to a halt and then begin the acceleration towards the starting point

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What force is required to cause a 5 kg bowling ball to accelerate at 4 m/s2​
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4 0
4 years ago
1. horizontal row in the periodic table____ 2. vertical column in the periodic table____ 3. A repetition of properties occurs wh
Anna11 [10]

Answer:

1. Periods

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Explanation:

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4 0
3 years ago
A truck moving at 6.0 m/s applies the brakes and stops after skidding 12.0 m.
Vitek1552 [10]

Answer:

a) a= -1.5 m/s²

b) t = 4.0 s

Explanation:

a)

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        v_{f} ^{2} - v_{o} ^{2} = 2*a*\Delta X  (1)

  • Since the truck finally stops, this means that vf = 0.
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       a =\frac{-(v_{o}^{2})}{2*\Delta X} = \frac{-(6.0m/s)^{2} }{2* 12.0m} = -1.5 m/s2  (2)

b)

  • Since we know the value of the initial and final velocity, and the value of the acceleration also, we can apply the definition of acceleration, solving for time t, as follows:

       \Delta t = \frac{-v_{o}}{a} = \frac{-6.0m/s}{-1.5m/s2} = 4.0 s  (2)

5 0
3 years ago
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