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Fittoniya [83]
2 years ago
15

Suppose that 2 J of work are needed to stretch a spring from its natural length of 30 cm to a length of 42 cm. How far (in cm) b

eyond its natural length will a force of 30 N keep this spring stretched
Physics
1 answer:
sergeinik [125]2 years ago
7 0

A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched

<u>Explanation:</u>

Work, W = 2 J

Initial distance, x_{1} = 30 cm

Final distance,  = 42 cm

Force, F = 30 N

Stretched length, x = ?

We know,

W = 1/2 kΔx²

Δx = 42-30 cm = 12 cm = 0.12 m

2 J = 1/2 k X (0.12)²

k = 277.77 N/m

According to Hooke's law,

F = kx

30 N = 277.77 X x

x = 0.108 m

x = 10.8  cm

A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched.

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yaroslaw [1]

First,

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where \rho is density, m is mass, and V is volume. We can compute the volume of the roll:

2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V

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When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness x. Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

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3 years ago
A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you
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Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

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v2: final speed = 0m/s (the balloon stops in my hands)

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The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

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