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allsm [11]
4 years ago
15

A 910-kg object is released from rest at an altitude of 1200 km above the north pole of the Earth. If we ignore atmospheric fric

tion, with what speed does the object strike the surface of the earth? (G = 6.67 × 10-11 N · m2/kg2, MEarth = 5.97 × 1024 kg, the polar radius of the earth is 6357 km)
Physics
1 answer:
gayaneshka [121]4 years ago
3 0

In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,

KE = \fract{1}{2}mv^2

PE = GMm(\frac{1}{r_1}-(\frac{1}{r_2}))

Where,

M = Mass of Earth

m = Mass of Object

v = Velocity

r = Radius

G = Gravitational universal constant

Our values are given as,

m = 910 Kg

r_1 = 1200 + 6371 km = 7571km

r_2 = 6371 km,

Replacing we have,

\frac{1}{2} mv^2 =  -GMm(\frac{1}{r_1}-\frac{1}{r_2})

v^2 =  -2GM(\frac{1}{r_1}-\frac{1}{r_2})

v^2 = -2*(6.673 *10^-11)(5.98 *10^24) (\frac{1}{(7.571 *10^6)} -\frac{1}{(6.371 *10^6)})

v = 4456 m/s

Therefore the speed of the object when striking the surface of earth is 4456 m/s

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Explanation:

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Answer:

The correct option is a

Explanation:

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A generator is designed to produce a maximum emf of 190 V while rotating with an angular speed of 3800 rpm. Each coil of the gen
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Answer:

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\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s

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The order of elements in the periodic table is based on
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Answer:

hope this helps

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