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Tju [1.3M]
3 years ago
9

We know that every object exerts an attraction on every other object and the heavier the object ___ the attraction.

Physics
1 answer:
hichkok12 [17]3 years ago
6 0
I don't know what the exact word is, but I do know that the bigger an objects mass is the more it will attract other objects toward it, mainly smaller objects with less mass. it might be gravity or something around those lines....is it a multiple choice question? 
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The momentum difference of a 3 kg mass with initial velocity of 8 m/s and a final velocity of 16 m/s
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Answer:

6

Explanation:

because if 3kg is 8m/s then 16 divided by 2 is 8 means add 3 to 3 and get the final answer

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Before you conduct your experiment, you need to form a hypothesis. A hypothesis is a prediction of what you think will happen in
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Example: A apple rotting.

If I put my apple in a fridge, then it would not rot as fast because it is in a cooled area. (example)

Hope it helps! Brainiest Answer would be amazing!

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3 years ago
Why do we learn useless nonsense school, why don't we learn what we will actually use in life
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I totally agree but, in my opinion its because of the government and what the state has control over. Teacher have little control over it.
4 0
3 years 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
Kitty [74]

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

= -1337.3 J

-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.

Learn more about friction work here:

brainly.com/question/14619763

#SPJ1

4 0
2 years ago
Tensile stress is: A. the strain per unit length. B. the ratio of elastic modulus to strain. C. the ratio of the change in lengt
victus00 [196]

Answer:

D. the same as force. the applied force per cross-sectional area.

Explanation:

Tensile stress of a material is defined as the ratio of the applied force on the material to its cross sectional area. this is expressed mathematically as;

Tensile stress = Force/cross sectional area

Tensile stress = F/A

Force is measured in newton while cross sectional area is measured in m

Hence the unit of Tensile stress is N/m²

6 0
3 years ago
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