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2 years ago

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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?

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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The speed vf of the object when a time t has passed is given by:

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Similarly, the distance y the object has traveled is calculated as follows:

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If we know the height h from which the object was dropped, we can solve the above equation for t:

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The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

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The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

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Therefore, it has traveled down a distance:

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Then, $K_2$ is equal to $K_1$ , so:

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<u>Answer:</u>

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We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

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We have equation of motion, $v^2=u^2+2as$ , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

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2 years ago

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marissa [1.9K]

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