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Lorico [155]
3 years ago
6

Newton's Law of Gravitation states that two bodies with masses m1 and m2 attract each other with a force F, where r is the dista

nce between the bodies and G is the gravitational constant.
F = G m1*m2/r^2
Use Newton's Law of Gravitation to compute the work W required to propel a 800 kg satellite out of the earth's gravitational field. You may assume that the earth's mass is 5.98 x 10^24 kg and is concentrated at its center. Take the radius of the earth to be 6.37 x 10^6 m and G = 6.67 x 10^-11 Nm^2/kg^2. (Round your answer to three significant digits.)
Physics
1 answer:
Marrrta [24]3 years ago
4 0

Answer:

W = - 5.01 10¹⁰ J

Explanation:

Work is defined by the expression

      W = ∫ F.dr

Where the blacks indicate vectors, in the case the force is radial and the distance is also radial, whereby the scalar producer is reduced to an ordinary product

      W = ∫ F dr

      W = G m₁m₂ ∫ 1 /r² dr

     W = G m₁ m₂2(-1 / r)

We evaluate between the lower limits r = Re and upper r = ∞

     W = G m₁m₂ (-1 / Re + 1 / ∞)

     W = - G m₁ m₂ / Re

Let's calculate

    W = - 6.67 10⁻¹¹ 800 5.98 10²⁴ / 6.37 10⁶

    W = - 5.01 10¹⁰ J

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A cat jumps from 2.5-meter-tall bookshelf to a 1.3-meter-tall countertop. If the cat
UkoKoshka [18]

The change in Potential energy of the cat is 176.4 J.

<h3 /><h3>Potential Energy:</h3>

This is the energy due to the position of a body. The S.I unit is Joules (J)

The formula for change in potential energy.

<h3 /><h3>Formula:</h3>
  • ΔP.E = mg(H-h).............. Equation 1

<h3>Where:</h3>
  • ΔP.E = Change in potential energy
  • m = mass of the cat
  • g = acceleration due to gravity
  • H = First height
  • h = second height.

From the question,

<h3>Given:</h3>
  • m = 15 kg
  • H = 2.5 m
  • h = 1.3 m
  • g = 9.8 m/s²

Substitute these values into equation 1

  • ΔP.E = 15×9.8(2.5-1.3)
  • ΔP.E = 15×9.8×1.2
  • ΔP.E = 176.4 J.

Hence, The change in Potential energy of the cat is 176.4 J

Learn more about Potential energy here: brainly.com/question/1242059

5 0
2 years ago
What actions can be explained by physics?
Vesna [10]
Всяко действие има равно по големина и противоположно по посока противодействие.
7 0
3 years ago
If you were stuck in the center of the bridge when it was swaying, where would be the safest place to walk back to land?
Ludmilka [50]
Straight
You already have to momentum of walking forward, and going back and forth are the same distance. If you go back then you would have to stop, turn and walk, but if you go forward you just have to walk.
4 0
3 years ago
A reconnaissance plane flies 605 km away from
kolezko [41]

Answer:

                      v_{avg}  = 355 m/s  

Explanation:

Distance = 605 km

Initial speed = v_{i} = 284 m/s

Final velocity = v_{f} = 426 m/s

Average speed = ?

There is two method two find average speed. In first method, using 3rd equation of motion, we find acceleration.

                        2as = v_{f}^{2}+v_{i}^{2}

Then using first equation of motion, we find time

                        v_{f} = v_{i}+at

Then using the formula of average velocity, we find average velocity

                         v_{avg}=\frac{total-distance}{total-time}

Second method is very simple

                                  v_{avg}=\frac{v_{f}+v_{i} }{2}

                                   v_{avg}=\frac{426+284}{2}

                                   v_{avg}  = 355 m/s      

8 0
4 years ago
A spring with spring constant of 34 N/m is stretched 0.12 m from its equilibrium position. How much work must be done to stretch
Nesterboy [21]

Answer:0.253Joules

Explanation:

First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.

F = ke where;

F is the force

k is spring constant = 34N/m

e is the extension = 0.12m

F = 34× 0.12 = 4.08N

To get work done,

Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.

Work done = Force × Distance

Since F = 4.08m, distance = 0.062m

Work done = 4.08 × 0.062

Work done = 0.253Joules

Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules

8 0
4 years ago
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