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3 years ago

How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to the "end of the universe"?

Physics

2 answers:

meriva3 years ago

6 0

Moon was rotation Earth or sun
it's own orbit . universe is not end

marysya [2.9K]3 years ago

3 0

Answer:

W(3R) - W(2R) = -PR² (1/(3R) - 1/(2R)) = PR/6

Explanation:

"Assume the weight move up at constant speed. With no net acceleration, the force applied is -(weight). Since the weight at height R is -P(R/x)² (minusbecause it's directed downward) the applied lifting force is P(R/x)², and the work done moving from x to x+dx is dW = P(R/x)² dx. Intetgrate this:"

W(x) = ?PR²/x² dx = -PR²/x + C

The work done moving from x=R to x=2R is:

W(2R) - W(R) = -PR²(1/(2R) - 1/R) = PR/2

(b) The work done moving from 2R to 3R is:

W(3R) - W(2R) = -PR² (1/(3R) - 1/(2R)) = PR/6

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Dmitry [639]

A pH scale goes from 0 - 14.
7 is in the middle so it is neutral

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3 years ago

Which bright solar feature is shown in the picture above?

Ghella [55]

Answer : (B) Prominence

Explanation :

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Photosphere is one of the layer of sun where the prominence are anchored and then they move into the corona of the sun.

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The prominence consists of colder plasma and this prominence plasma is much more shining and denser as compared to coronal plasma.

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6 0

2 years ago

Read 2 more answers

IF YOU GOOD AT SCIENCE PLEASE ANSWER THIS ASAP

xz_007 [3.2K]

Answer:

I would believe that it would be the last option

Explanation:

Physical science is a type of science that mainly focuses on natural objects that are not alive, such as minerals and rocks.

4 0

1 year ago

Read 2 more answers

A coin released at rest from the top of a tower hits the ground after falling 5.7 s. What is the speed of the coin as it hits th

Bad White [126]

Given parameters:

Initial velocity of Coin = 0m/s

Time taken before coin hits ground = 5.7s

Unknown:

Final velocity of the coin = ?

Velocity is displacement with time. To solve this problem, we have to apply one of the equations of motion.

The fitting one of them here is shown below;

V = U + gt

where;

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

t is the time taken

Here we use positive value of acceleration due to gravity because the coin is falling with the effect of acceleration and not against it.

Now input the parameters and solve;

V = 0 + 9.81 x 5.7

V = 55.917m/s

Therefore, the final velocity is 55.917m/s.

8 0

2 years ago

A player is positioned 35 m[40 degrees W of S] of the net. He shoot the puck 25 m [E] to a teammate. What second displacement do

Lubov Fominskaja [6]

Answer:

x=22.57 m

Explanation:

Given that

35 m in W of S

angle = 40 degrees

25 m in east

From the diagram

The angle

$\theta=90-40=50^o$

From the triangle OAB

$cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}$

$1340.57=35^2+25^2-x^2$

x=22.57 m

Therefore the answer of the above problem will be 22.57 m

4 0

2 years ago