Calculating Intensity of Spherical Wave & Pendulum T - Confirm Correct Answers

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The intensity of a spherical wave at a distance of 3 m from the source is calculated to be approximately 588 W/m^2 using the inverse square law. The formula I_2/I_1 = (r_1)^2/(r_2)^2 confirms that intensity decreases with the square of the distance from the source. For the pendulum with an original period of 10 seconds, halving the length results in a new period of about 7.1 seconds. The relationship between the pendulum's period and its length is derived from T = 2π√(L/g), confirming the calculations. Both answers for intensity and pendulum period are validated as correct.
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The intensity of a spherical wave is 21 W/m^2 at a distance 16 m away from a constantly emitting source. What will it be 3 m away?

Using the formula

I_2/I_1 = (r_1)^2/(r_2)^2

I got 588 W/m^2 0r 600. Correct?


If the length of pendulum with T = 10 s were halved, what would be its new T?

My answer was 7.1. Correct?

Halved scenario:

T = 2pi*sqrt(L/2g)

but first find original L with

10 s = 2pi*sqrt(L/9.8) to plug in.

Thanks
 
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first answer is correct: I is proportional to the inverse square since the source emits energy in spherical wave fronts and the surface area of a sphere is 4*pi*r^2.

the second anser sounds good but i haven't a calculator near me. The period of a pendulum is proportional to the square root of the length so T1/T2=(L1/L2)^.5 so the answer reduces to 10/2^.5 or 10/1.4
 
The second is right too.
 
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