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Eddi Din [679]
3 years ago
5

(20 POINTS!!!) An object that is 0.5 m above the ground has the same amount of potential energy as a spring that is stretched 0.

5 m. Each distance is then doubled. How will the potential energies of the object and the spring compare after the distances are doubled? The gravitational potential energy of the object will be two times greater than the elastic potential energy of the spring. The elastic potential energy of the spring will be four times greater than the gravitational potential energy of the object. The elastic potential energy of the spring will be two times greater than the gravitational potential energy of the object. The potential energies will remain equal to one another.
Physics
2 answers:
kompoz [17]3 years ago
8 0

The correct answer is The elastic potential energy of the spring will be two times greater than the gravitational potential energy of the object.

The formula for Gravitational potential energy is= mgh where

m=mass

g= 9.8

h=height.

On the other hand the formula for Elastic potential energy is (1/2)KX^2

where K is the spring constant and x is the displacement of the string. By changing the values of H and X, we will see elastic potential energy will remain more.

Andrei [34K]3 years ago
4 0
C is the answer. <span>The elastic potential energy of the spring will be two times greater than the gravitational potential energy of the object.</span>
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Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
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Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

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Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

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3 years ago
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