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egoroff_w [7]
3 years ago
14

(a) Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass mE, radius RE) and

place it in a circular low earth orbit, that is, an orbit whose altitude above the earth's surface is much less than RE. (As an example, the International Space Station is in low earth orbit at an altitude of about 400km, much less than RE = 6370km .) Ignore the kinetic energy that the spacecraft has on the ground due to the earth’s rotation.
(b) Calculate the minimum amount of additional work required to move the space craft from low earth orbit to a very great distance from the earth. Ignore the gravitational effects of the sun, the moon, and the other planets.
(c) Justify the statement "In terms of energy, low earth orbit is halfway to the edge of the universe.

Physics
1 answer:
notsponge [240]3 years ago
8 0

Answer:

(a) The Work need to   place spacecraft into low orbit isGmm_{E} /2R_{E}

(b) Additional Work need to  place the spacecraft far away from the earth is

Gmm_{E} /2R_{E}.

(c) From part(a) and part(b) we observe the Both quantities are equal.

Explanation:

Explanation is in the following attachments

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Answer:

Final speed of the box after it has moved to x = 10 is given as

v = 3.35 m/s

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W_{ex} + W_{fric} = \frac{1}{2}mv^2

F x - \mu mg x = \frac{1]{2}mv^2

16.27 (10) - (0.15)(8)(9.8) (10) = \frac{1}{2}(8) v^2

162.7 - 117.6 = 4 v^2

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3 years ago
How does electricity flow?
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According to coulombs law, what will happen to the force between two charged particles if the magnitude are increased by 6 times
seropon [69]

Answer: The electrostatic force will be the same

Explanation:

According to Coulomb's Law, when two electrically charged bodies come closer, appears a force that attracts or repels them, depending on the sign of the charges of this two bodies or particles.  

In this sense, this law states the following:

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"  

 

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}  (1)

Being K is a proportionality constant.  

Now, if each q_{1} and q_{2} are increased by 6, and the distance between them as well, we will have the following:

F_{E}= K\frac{6 q_{1}. 6 q_{2}}{(6d)^{2}}  (2)

F_{E}= 36 K\frac{q_{1}. q_{2}}{36d^{2}}  (3)

Simplifying:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}  (4)

Comparing (1) with (4) we can see the electrostatic force is the same.

4 0
4 years ago
Express each of the following in ms -1 <br>a) 18kmh-1<br>​
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5ms-1
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3 years ago
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