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3 years ago

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Hooke's law. The distance d when a spring is stretched by a hanging objective varies directly as the weight w of the object. If

the distance is 38 cm when the weight is 3 kg, what is the distance when the weight is 7 kg? ...?

1 answer:

Fofino [41]3 years ago

7 0

<span>Hooke's law is F=-kx, which means the elastic force contained by the spring is a product of the distance it stretches and its spring constant, but the direction of the force is opposite that of the displacement. We calculate as follows:

</span><span>(3 kg)(9.8 m/s^2) = -k(-0.38 m)
</span>k =<span> 77.4

</span><span>Then use k to find the new displacement, again using Hooke's law:
(7 kg)(9.8 m/s^2) = -(77.4)x
x = -0.89 m</span>

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Answer:

Part a)

$H = 44.1 m$

Part b)

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Part c)

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Explanation:

Part a)

As we know that ball will reach at maximum height at

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$H = 44.1 m$

Part b)

Height of the fence is given as

$y = (vsin\theta) t - \frac{1}{2}gt^2$

$y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)$

$y = 13.48 m$

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

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$97.5 = v_x (5.5)$

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Explanation:

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The minus sign in the magnification is because the image is inverted.

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