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Musya8 [376]
2 years ago
11

A 250 Hz tuning fork is struck and the intensity at the source is I1 at a distance of one meter from the source. (a) What is the

intensity at a distance of 4.00 m from the source? (b) How far from the tuning fork is the intensity a tenth of the intensity at the source?
Physics
1 answer:
Zina [86]2 years ago
3 0

Answer:

a) 0.0625 I_1

b) 3.16 m

Explanation:

<u>Concepts and Principles  </u>

The intensity at a distance r from a point source that emits waves of power P is given as:  

I=P/4π*r^2                         (1)

<u>Given Data</u>

f (frequency of the tuning fork) = 250 Hz

I_1 is the intensity at the source a distance r_1 = I m from the source.  

<u>Required Data</u>

- In part (a), we are asked to determine the intensity I_2 a distance r_2 = 4 in from the source.

- In part (b), we are asked to determine the distance from the tuning fork at which the intensity is a tenth of the intensity at the source.  

<u>solution:</u>

(a)  

According to Equation (1), the intensity a distance r is inversely proportional to the distance from the source squared:

I∝1/r^2

Set the proportionality:  

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

Solve for I_2 :  

I_2=I_1(r_2/r_1)^2  

I_2=0.0625 I_1

(b)  

Solve Equation (2) for r_2:  

r_2=(√I_1/I_2)*r_1

where I_2 = (1/10)*I_1:

r_2=(√I_1/1/10*I_1)*r_1

     =3.16 m

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as well.

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8 0
1 year ago
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What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade?
kaheart [24]

Incomplete question.The complete question is here

What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade? The outer edge of the blade is 21 cm from the center of the blade, and the mass of the outer portion is 7.7 g. Even though the blade is 21cm long, the last 2cm should be treated as if they were at a point 20cm from the center of rotation.

Answer:

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Explanation:

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F= m×ω²×r

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