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zmey [24]
3 years ago
13

A spacecraft is separated into two parts by detonating the explosive bolts that hold them together. The masses of the parts are

1150 kg and 1900 kg. The magnitude of the impulse on each part from the bolts is 200 N·s. With what relative speed do the two parts separate because of the detonation?
Physics
1 answer:
nordsb [41]3 years ago
6 0

Answer:

  • 0.279 m/s

Explanation:

mass (m1) = 1150 kg

mass (m2) = 1900 kg

impulse (J) = 200 N.s

With what relative speed do the two parts separate because of the detonation?

since both parts separate, their relative speed = speed of m1 + speed of m2

  • speed of m1 = \frac{J}{m} = \frac{200}{1150} = 0.174 m/s
  • speed of m2 = \frac{J}{m} = \frac{200}{1900} = 0.105 m/s
  • relative speed = 0.174 + 0.105 = 0.279 m/s

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Answer:

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3 years ago
Two equipotential surfaces surround a +3.10 x 10-8-c point charge. how far is the 290-v surface from the 41.0-v surface?
MrMuchimi
 T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r 

where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance. 
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