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miv72 [106K]
3 years ago
8

How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to an altitude of 5r miles above t

he surface of the moon (i.e., x = 6r)?
Physics
1 answer:
Cloud [144]3 years ago
7 0

Let the data is as following

mass of payload = "m"

mass of Moon = "M"

now we know that we place the payload from the position on the surface of moon to the position of 5r from the surface

So in this case we can say that change in the gravitational potential energy is equal to the work done to move the mass from one position to other

so it is given by

W = U_f - U_i

we know that

U_f = -\frac{GMm}{6r}

U_i = -\frac{GMm}{r}

now from above formula

W = -\frac{GMm}{6r} + \frac{GMm}{r}

W = \frac{5GMm}{6r}

so above is the work done to move the mass from surface to given altitude

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PLEASE HELP ILL MARK BRAINLIEST!!! A ball is initially thrown downwards with an initial speed of 20 m/s from the top of a 300 m
Sholpan [36]

Using the 3rd equation of motion:

= v² - u² = 2gs ------ [g = Acceleration due to gravity]

= v² - 20² = 2 × 10 × 300

= v² - 400 = 6000

= v² = 6000 - 400

= v = √5600

= v = 74.83 m/s

And yeah it's done :)

8 0
3 years ago
If two objects are moving at the same speed, and Object 1 has four times as much mass as Object 2, how much momentum does Object
Margaret [11]
The answer is  number 1
7 0
3 years ago
The total distance a group of movers move a refrigerator is 22 m. If they complete this task in 15 minutes, their efforts would
lianna [129]

Answer:

Their efforts would be expressed in units of Joules per second

Explanation:

The unit of their efforts can be derived from the formula of power which is given by the product of mass, acceleration and distance (the product is energy with unit joules) divided by time taken to complete the task (unit is seconds)

Therefore, the unit of their efforts would be joules per second

4 0
3 years ago
A car of mass 750 kg accelerates away from traffic lights. At the end of the first 100 m it has reached a speed
malfutka [58]

the work done on the car by the force of its engine is 78,000 J.

" The work done on the car by the force of friction is 24,000 J.

Increasing the car's kinetic energy at the end of the first 100 m is 54,000J

a. Completed work = force x distance. Engine output = 780 N, that is,

780 N x 100 m = 78,000 J.

b. Completed work = force x distance. Friction force = 240 N, that is,

240 N x 100 m = 24,000 J.

c. Kinetic energy = 1 \ 2 x m x v2

= 1 \ 2 x 750 kg x 12 squared = 375 x 144 = 54,000 J.

<h3>How powerful is the engine of a car? </h3>

Mainstream car and truck engines typically produce 100-400 pounds. -Torque feet. This torque is generated by the engine piston as it moves up and down on the engine crankshaft, causing the engine to rotate (or twist) continuously.

Learn more about work done here:  brainly.com/question/25573309

#SPJ10

5 0
2 years ago
Which of the following statements is not one of Newton's laws of motion?
Alex17521 [72]

Answer:

The answer to your question: d.

Explanation:

a. The rate of change of momentum of an object is equal to the net force applied to the object.  

This is the second a law of motion, so this answer is incorrect.

b. In the absence of a net force acting on it, an object moves with constant velocity.

This is the first Newton law of motion, so this option is not correct.

c. For any force, there always is an equal and opposite reaction force.

This is the third law of motion, so this is not the right option.

d. What goes up must come down.

Newton said this sentence, but is not part of the law of motion.

5 0
3 years ago
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