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MissTica
2 years ago
5

The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc

ope is 9.21 × 104 N, and the mass of Earth is 5.98 × 1024 kg, what is the mass of the telescope? Round the answer to the nearest whole number
Physics
2 answers:
Luda [366]2 years ago
5 0
The answer is 11,121 kg
PtichkaEL [24]2 years ago
4 0

The gravitational attraction between the Earth and the Hubble telescope is given by:

F=G\frac{M m}{r^2}

where

G is the gravitational constant

M is the Earth's mass

m is the Hubble mass

r is the distance of the Hubble telescope from the earth's center


By rearranging the equation and substituting the numbers given by the problem, we find the mass of the telescope:

m=\frac{F r^2}{GM}=\frac{(9.21 \cdot 10^4 N)(6.94 \cdot 10^6 m)^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(5.98 \cdot 10^{24}kg) }  = 11121 kg

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Replacing by the values, and solving for vf, we get:

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If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

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So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

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What happens when calcium reacts with chlorine?
larisa86 [58]

Answer:

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