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VladimirAG [237]
3 years ago
6

The gravitational force, F, between an object and the Earth is inversely proportional to the square of the distance from the obj

ect and the center of the Earth. If an astronaut weighs 209 pounds on the surface of the Earth, what will this astronaut weigh 2450 miles above the Earth?
Physics
1 answer:
andreev551 [17]3 years ago
3 0
<h2>Weight of astronaut 2450 miles above the Earth is 80.38 pounds</h2>

Explanation:

Given that gravitational force, F, between an object and the Earth is inversely proportional to the square of the distance from the object and the center of the Earth.

             F=\frac{k}{r^2}

Where F is gravitational force  between an object and the Earth, r is the distance from the object and the center of the Earth and k is a constant.

Radius of Earth = 4000 miles

In case 1 an astronaut weighs 209 pounds on the surface of the Earth,

              209=\frac{k}{4000^2}\\\\k=3.344\times 10^9

Now we need to find weight of astronaut 2450 miles above the Earth    

              r = 4000 + 2450 = 6450 miles

               F=\frac{k}{6450^2}\\\\F=\frac{3.344\times 10^9}{6450^2}=80.38pounds

Weight of astronaut 2450 miles above the Earth is 80.38 pounds  

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Answer:

The speed of the vehicles immediately after the collision is 5.84 m/s.

Explanation:

The speed of the vehicles after the collision can be found by conservation of linear momentum:

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Where:

m₁: is the mass of the car = 0.5 ton = 500 kg

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v_{2_{f}}: is the final speed of the lorry =?    

Since the two vehicles become tightly locked together after the collision v_{1_{f}} = v_{2_{f}}:

m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = v(m_{1} + m_{2})

v = \frac{m_{1}v_{1_{i}} + m_{2}v_{2_{i}}}{m_{1} + m_{2}} = \frac{500 kg*11.11 m/s + 9500 kg*5.56 m/s}{500 kg + 9500 kg} = 5.84 m/s

Therefore, the speed of the vehicles immediately after the collision is 5.84 m/s.

I hope it helps you!  

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Answer:

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