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tia_tia [17]
3 years ago
5

Which paragraph shows the correct relationship between kinetic energy and speed?

Physics
2 answers:
Troyanec [42]3 years ago
8 0

Answer:

Graph?

Explanation:

Mademuasel [1]3 years ago
5 0

I cant see the paragraph so i cant see. It srry

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Gravity on Jupiter is 25 m/s/s, what is the weight of a 12 kg object on Jupiter?
stiv31 [10]

Answer:

300 Newtons

Explanation:

Weight is the force of attraction between two bodies, one usually larger (like a planet), and one smaller (like a person). Force can be calculated using the formula: Force = mass × acceleration.

The mass here is 12kg, the acceleration, which in this case, is the acceleration due to gravity is 25m/s/s, by plugging in our values, we have

Force = 12 × 25 = 300 Newtons or 300 N for short.

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When a pendulum is at the position all the way to the left when it is swinging (at the top of the arc), what is true of the kine
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Potential energy + kinetic energy = constant at every moment in time

At the highest point:

potential energy is at its maximum 
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Which statement best explains the movement of electric current from the clouds to the ground during a lightning storm? .
olga2289 [7]
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If a baseball is in space and someone hits it with a bat what does the ball do
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It's will fly like float

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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
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