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madreJ [45]
3 years ago
9

An athlete stretches a spring an extra 40.0 cm beyond its initial length. How much energy has he transferred to the spring, if t

he spring constant is 52.9 N/cm?
a) 4230 kJ
b) 4230 J
c) 423 kJ
d) 423 J
Physics
1 answer:
notsponge [240]3 years ago
4 0

Answer:

<h2>b) 4230 J </h2>

Explanation:

Step one:

given data

extension= 40cm

Spring constant K= 52.9N/cm

Step two:

Required

the Kinetic Energy KE

the expression to find the kinetic energy is

KE= 1/2ke^2

substituting our data we have

KE= 1/2*52.9*40^2

KE=0.5*52.9*1600

KE= 42320Joules

<u>The answer is b) 4230 J </u>

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Answer:

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Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

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We calculate the range

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We will calculate the time of flight,

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When the ball reaches the end point has the same initial  height Y=0

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We solve the equation

0= t ( 12.5 -0.815 t)

 t=0 s

t= 15.3 s

The value of zero corresponds to the departure point and the flight time is 15.3 s

Let's calculate the reach on earth

R2 = 25² sin (2 30) / 9.8

R2 = 55.2 m

R/R2 = 332/55.2

R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

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