How high does a rocket have to go above the Earth’s surface so that its weight is reduced to 58.8 % of its weight at the Earth’s
surface? The radius of the Earth is 6380 km and the universal gravitational constant is 6.67 × 10−11 N · m 2 /kg2 . Answer in units of km
1 answer:
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Answer:
70 cm
Explanation:
0.5 kg at 20 cm
0.3 kg at 60 cm
x = Distance of the third 0.6 kg mass
Meter stick hanging at 50 cm
Torque about the support point is given by (torque is conserved)
The position of the third mass of 0.6 kg is at 20+50 = 70 cm