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Pachacha [2.7K]
3 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]3 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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What is the potential energy of the system composed of the three charges q1, q3, and q4, when q1 is at point R? Define the poten
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Answer:

Incomplete question check attachment for complete question

Also it is given that

q1=-0.7uC

q4=-1.7uC

q3=-1.7uC

Also the distance are given as

a=2.2cm=0.022m

d2=3.6cm=0.036m

Explanation:

The potential energy due to point R is given as

The potential energy due to charge q1 and q3 plus the potential energy due to charge q4 and q1 plus the potential energy due to charge q3 and q4

So, let take it one after the other

Potential energy is give as

P.E=kq1q2/r

Therefore,

Potential energy due to charge q1 and q3

U¹³=kq1q3/r

To get the distance between charge q1 and q3, we will apply Pythagoras theorem

r=√(d2²+a²)

r=√(0.036²+0.022²)

r=0.0422m

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Then,

U¹³=kq1q3/r

U¹³=9×10^9×0.7×10^-6×1.7×10^-6/0.0422

U¹³=0.254J

Potential energy due to charge q1 and q4

U¹⁴=kq1q4/r

To get the distance between charge q1 and q4, we will apply Pythagoras theorem

r=√(d2²+a²)

r=√(0.036²+0.022²)

r=0.0422m

k is a constant =9×10^9Nm²/C²

Then,

U¹⁴=kq1q4/r

U¹⁴=9×10^9×0.7×10^-6×1.7×10^-6/0.0422

U¹⁴=0.254J

Potential energy due to charge q3 and q4

U³⁴=kq3q4/r

r=2a=2×0.022=0.044m

k is a constant =9×10^9Nm²/C²

Then,

U³⁴=kq3q4/r

U³⁴=9×10^9×1.7×10^-6×1.7×10^-6/0.044

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Then, the total energy is

U= U¹³+ U¹⁴ + U³⁴

U=0.254+0.254+0.591

U=1.099J

Then also, the potential energy is zero because at infinity both U¹³ and U¹⁴ will have infinite potential because their distance apart will be infinite.

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Explanation:

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