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Arturiano [62]
2 years ago
14

A spring of spring constant k=8.25N/m is displaced from equilibrium by a distance of 0.150m. What is the stored energy in the fo

rm of spring potential energy? Show your work.
Physics
1 answer:
inysia [295]2 years ago
4 0

Answer:

0.0928J

Explanation:

the pulling force of spring F=-kx

where x is the displacement from equilibrium position.

energy stored:

\int\limits^x_0 {-F} \, dx \\=\int\limits^x_0 {kx} \, dx \\\\=\frac{kx^{2} }{2}

*** Its fine if you know nothing about calculus. Just apply the equation

    U=\frac{kx^{2} }{2}

  where U is the potential energy of the spring***

  put x=0.150, U=\frac{8.25}{2}×0.150^{2} = 0.0928J (corr. to 3 sig. fig.)

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Jeremy has a handful of resistors. All of them, except one, have four bands. The exception has three. Why? A. He should use anot
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A stone is dropped from a cliff at time t = 0 and strikes the ground below.
Dafna1 [17]

Answer:

S=

2

1

gt

2

........(1)

And that of the other stone is

S=u(t−n)+

2

1

[g(t−n)

2

]........(2)

Since both the stones meet at the distance so equation (1)and equation(2) will be equal

2

1

gt

2

=u(t−n)+

2

1

[g(t−n)

2

]

gt

2

=2ut−2un+gt

2

+gn

2

−2gnt

t(2gn−2u)=gn

2

−2un

t=

(gn−u)

n(

2

gn

−u)

now putting value of t in equation(1)

s=

2

g

⎣

⎢

⎡

gn−u

2

[n(

2

gn

−u)]

⎦

⎥

⎤

2

S=

2

g

⎣

⎢

⎡

(gn−u)

2

n(

2

gn

−u)

⎦

⎥

⎤

Explanation:

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