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o-na [289]
2 years ago
9

A block of mass m is pushed up against a spring with spring constant k until the spring has been compressed a distance x from eq

uilibrium. what is the work done on the block by the spring.
A. W=kx^2
B.W=-kx^2
C.W=0
D. None of these
Physics
1 answer:
Snowcat [4.5K]2 years ago
6 0

Answer:d

Explanation:

Spring is compressed to a distance of x from its equilibrium position

Work done by block on the spring is equal to change in elastic potential energy

i.e. Work done by block W=\frac{1}{2}kx^2

therefore spring will also done an equal opposite amount of work on the block in the absence of external force

Thus work done by spring on the block W=-\frac{1}{2}kx^2

Thus option d is correct

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Answer:

Where is the text?

Explanation:

If you refer to the short sentence you wrote as text, I believe the answer is probably the word "crashes" because it shows how the momentum was transferred.

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Two workhorses tow a barge along a straight canal. Each horse exerts a constant force of magnitude F, and the tow ropes make an
morpeh [17]

Answer:

<em>a) Fvt cosθ</em>

<em>b) Fv cosθ</em>

<em></em>

Explanation:

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<em></em>

b) Power provided by the horse P = force x speed

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2 years ago
Pamela drove her car 999999 kilometers and used 999 liters of fuel. she wants to know how many kilometers (k)(k)left parenthesis
Vanyuwa [196]
When the relationship between two variables are said to be proportional, it means that one variable is a constant multiple of the other variable. They are related by a constant of proportionality, usually denoted as k. 

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distance = k * liters of fuel, such that 
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3 0
2 years ago
A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the
hoa [83]

Answer:

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'A' is the amplitude = 6 inches (given)

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At equilibrium we have

mg=kx\\\\k=\frac{mg}{x}

Applying values we get

k=40 lb/ft

thus natural frequency equals

\omega =\sqrt{\frac{40}{\frac{20}{32}}}\\\\\omega =8s^{-1}

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x(t)=6sin(8t+\phi)

At time t=0 since mass is at it's maximum position thus we have

A=Asin(\omega t+\phi)\\\\\therefore sin(\omega\times 0+\phi)=1\\\\\phi=\frac{\pi}{2}\\\\\therefore x(t)=Asin(\omega t+\frac{\pi}{2})

Thus the position of mass at the given times is as follows

1) at \frac{\pi}{12} x(t)=5.99inches

2) at \frac{\pi}{8} x(t)=5.9909inches

3) at \frac{\pi}{6} x(t)=5.98397inches

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Answer:

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Explanation:

i want the answer i don't know

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