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Pachacha [2.7K]
2 years ago
9

A spring stretches from 10cm to 22cm when a force of 4n is applied, if it obeys Hooke’s law,its total length in cm when a force

of 6N is applied is?
Physics
1 answer:
satela [25.4K]2 years ago
7 0

Answer:

Explanation:

Hooke's Law is a linear representation that, in equation form, looks like this:

F = -kΔx where F is force, k is the spring constant, and Δx is the displacement of the spring when a Force is applied to it in the form of a mass hanging on the end of it. We need to use the equation to solve for the spring constant, k. Doing this by filling in the values we were given:

4 = -k(-12) the 12 is negative because the mass hangs below the point of equilibrium. This gives us that

k = 1/3. Now we'll use that in the next equation where we need to solve for displacement:

6 = -1/3(Δx) and

Δx = -18 cm

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Answer:

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Explanation:

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2 years ago
A jewellery melts 500g of Silver to pour into a mould. Calculate how much energy was released as the silver solidified.
irga5000 [103]

When silver is poured into the mould the it will solidify

In this process the phase of the Silver block will change from liquid to solid.

This phase change will lead to release in heat and this heat is known as latent heat of fusion.

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Answer:

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Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

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