How far will 20 N of force stretch a spring with a spring constant of 140 N/m?
1 answer:
Answer:
0.143 m
Explanation:
The relationship between force applied on a string and stretching of the spring is given by Hooke's law:
where
F is the force exerted on the spring
k is the spring constant of the spring
x is the stretching of the spring from its equilibrium position
In this problem, we have:
F = 20 N is the force applied on the spring
k = 140 N/m is the spring constant
Solving for x, we find how far the spring will stretch:
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Answer:
<em>6.02 dB increase </em>
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Explanation:
Let us take the initial power from the speaker P' = P Watt
then, the final power P = 4P Watt
for a given unit area, initial intensity (power per unit area) will be
I' = P Watt/m^2
and the final quadrupled sound will produce a sound intensity of
I = 4P Watt/m^2
Increase in loudness is gotten from the relation
ΔL =
where
I = final sound intensity
I' = initial sound intensity
imputing values of the intensity into the equation, we have
==> =
= <em>6.02 dB increase </em>