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eduard
3 years ago
15

what type of energy appears when a gymnast jumps on to a spring board? Question is on energy stores!Thanks

Physics
1 answer:
Brrunno [24]3 years ago
5 0
Hello There!

It is Spring potential energy. Also called Elastic potential energy.

Hope This Helps You!
Good Luck :) 

- Hannah ❤
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A 25N force is acting on a body moving on a straight line with Initial momentum 20 kam's. Find the final momentum after 4 second
Nezavi [6.7K]

The final momentum of the body is equal to 120 Kg.m/s.

<h3>What is momentum?</h3>

Momentum can be described as the multiplication of the mass and velocity of an object. Momentum is a vector quantity as it carries magnitude and direction.

If m is an object's mass and v is its velocity then the object's momentum p is: {\displaystyle \mathbf {p} =m\mathbf {v} . The S.I. unit of measurement of momentum is kg⋅m/s, which is equivalent to the N.s.

Given the initial momentum of the body = Pi = 20 Kg.m/s

The force acting on the body, Pf = 25 N

The time, Δt = 4-0 = 4s

The Force is equal to the change in momentum: F ×Δt = ΔP

25 × 4 = P - 20

100 = P - 20

P = 100 + 20 = 120  Kg.m/s

Therefore, the final momentum of a body is 120 Kg.m/s.

Learn more about momentum, here:

brainly.com/question/4956182

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5 0
1 year ago
The dimensions of the disk of our milky way galaxy are:.
Katen [24]

Answer:

According to studies, the milky way is approximately, "170,000–200,000 light-years (52–61 kpc) in diameter and, on average, approximately 1,000 ly (0.3 kpc) thick."

With that being said, it is safe to say that the dimensions are somewhere around 100,000 by 1,000

6 0
3 years ago
Question 2 of 5
Andrews [41]
Lifting weights, did this yesterday :))
4 0
3 years ago
Read 2 more answers
A bicycle wheel with a radius of 0.42 m accelerates uniformly for 6.8 s from an initial angular velocity of 5.5 rad/s to a final
Art [367]

Answer:

The angular acceleration required  is 0.1765 rad/ s^2

Explanation:

The radius of the bicycle wheel has a radius of 0.42 m.

The acceleration is for time, t =  6.8 seconds.

Initial angular velocity is given as  \omega_{0}  = 5.5 rad/s

Final angular velocity is given as \omega_{f} = 6.7 rad/s

Therefore from the formula for angular speed we get

\omega_{f} = \omega_{0} + (\frac{d\omega}{dt} \times t),   where t is the time in seconds.

Therefore we get

6.7 =  5.5 + (6.8 × \frac{d\omega}{dt} )

Therefore we get the angular acceleration, \frac{d\omega}{dt} = \frac{(6.7 - 5.5 }{6.8}  = 0.1765 rad/ s^2

The angular acceleration required  is 0.1765 rad/ s^2

8 0
3 years ago
What disagrees with laplace hypothesis of the suns formation
slega [8]

Answer:

The sun's mass moved toward the outer edge of it

Explanation:

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3 years ago
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