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ArbitrLikvidat [17]
3 years ago
15

Falling Faster

Physics
1 answer:
Daniel [21]3 years ago
3 0
In a vacuum they would fall at the same speed, but since it is only off the empire State building, the hammer would fall first

Gravity
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A spaceship maneuvering near planet zeta is located at r⃗ =(600i^−400j^+200k^)×103km, relative to the planet, and traveling at v
slava [35]

<u>Answer:</u>

 The spaceship's position when the engine shuts off = (708.15 i - 444.1 j + 200 k)*10^3km

<u>Explanation:</u>

  Initial location of spaceship = (600 i - 400 j + 200 k)*10^3km= (600 i - 400 j + 200 k)*10^6m

  Initial velocity = 9500 i m/s

  Acceleration = (40 i - 20 k)10^3m/s^2

  Time = 35 minute = 35 * 60 = 2100 seconds

 We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  Substituting

       s= (9500 i)*2100+\frac{1}{2}*(40 i - 20 k)*2100^2\\ \\ s=9500*2100 i+20*2100^2i-10*2100^2j\\ \\ s=19.95*10^6i+88.2*10^6i-44.1*10^6j\\ \\ \\s=(108.15i-44.1j)*10^6m

     So final position = ((600 i - 400 j + 200 k)+(108.15i-44.1j))*10^6=(708.15 i - 444.1 j + 200 k)*10^6m

                              =(708.15 i - 444.1 j + 200 k)*10^3km

    The spaceship's position when the engine shuts off = (708.15 i - 444.1 j + 200 k)*10^3km

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Explanation:

Hope this helps :D

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If one cup falls down then there will be 59 cups left.

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