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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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We can solve this part by using Newton's second law:

F=ma (1)

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m = 1177.5 kg

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2) 161.0 m

We can solve this part by using the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance travelled

u is the initial velocity

t is the time

a is the acceleration

Here we have

u = 0 (the boat starts from rest)

a=1.028 m/s^2

Substituting t = 17.7 s, we find the distance covered:

s=0+\frac{1}{2}(1.028)(17.7)^2=161.0 m

3) 18.2 m/s

The speed of the boat can be found with the following suvat equation

v=u+at

where

v is the final velocity

u is the initial velocity

t is the time

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u = 0 (the boat starts from rest)

a=1.028 m/s^2

And substituting t = 17.7 s, we find the final velocity:

v=0+(1.028)(17.7)=18.2 m/s

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Klio2033 [76]

Answer:

40·919 m

Explanation:

Initial velocity of the arrow = 46 m/s

Angle at which it is thrown from horizontal = 38°

<h3>At the maximum height, the vertical component of velocity will be 0</h3>

Initial velocity in vertical direction = 46 × sin(38) = 28·32 m/s

From the formula

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