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murzikaleks [220]
3 years ago
6

Forces can act on an object in the same direction or in opposite. how does each situation affect the motion of the object?

Physics
1 answer:
Anon25 [30]3 years ago
5 0



Hi pupil here's your answer ::




➡➡➡➡➡➡➡➡➡➡➡➡➡



Action and Reaction do not act on the same body !! If they acted on the same body, the resultant force will be zero and their could be never accelerated motion.

If both the forces acted on the same body, then if they are equal to opposite direction the object will remain stationary. If on of the forces is greater than other the object will move in the direction of greater force.

If both acted in the same direction there would be an accelrated motion.




⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅





Hope this helps . . . . .
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Answer:

22Km/sec

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When you drop a object on earth, earth gets pulled slightly toward that object as well. How do I calculate the force if the eart
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The gravitation force with which the earth is being pulled can be determined by applying Newton's law of universal gravitation.

<h3>What is gravitation force?</h3>

According Newton's law of universal gravitation, the force exerted between two objects in the universe is directly proportional to the product of masses of the two objects and inversely proportional to the square of the distance between the two objects.

Mathematically, the formula for gravitation force is given as;

F = GmM/R²

where;

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  • R is the distance of the object from earth
  • G is universal gravitation constant

If the mass of the object is know and the distance between earth  and the object is also known, the force with which the earth is being pulled can be calculated by applying Newton's law of universal gravitation as shown in the above equation.

Thus, the force with which the earth is being pulled can calculated as well.

Learn more about gravitation force here: brainly.com/question/27943482

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1 year ago
Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
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a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

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-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

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inysia [295]
The independent variable would be the cleaning products. 

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