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2 years ago

BEST ANSWER GETS BRAINLIEST!

Physics

1 answer:

wel2 years ago

8 0

Answer:

Distance = 16.9 m

Explanation:

We are given;

Power; P = 70 W

Intensity; I = 0.0195 W/m²

Now, for a spherical sound wave, the intensity in the radial direction is expressed as a function of distance r from the center of the sphere and is given by the expression;

I = Power/Unit area = P/(4πr²)

where;

P is the sound power

r is the distance.

Thus;

Making r the subject, we have;

r² = P/4πI

r = √(P/4πI)

r = √(70/(4π*0.0195))

r = √285.6627

r = 16.9 m

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You set a tuning fork into vibration at a frequency of 723 Hz and then drop it off the roof of the Physics building where the ac

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Answer:

Explanation:

Given

Original Frequency $f=723\ Hz$

apparent Frequency $f'=697\ Hz$

There is change in frequency whenever source move relative to the observer.

From Doppler effect we can write as

$f'=f\cdot \frac{v-v_o}{v+v_s}$

where

$f'=$ apparent frequency

v=velocity of sound in the given media

$v_s=$ velocity of source

$v_0=$ velocity of observer

here $v_0=0$

$697=723\cdot (\frac{343-0}{343+v_s})$

$v_s=(\frac{f}{f'}-1)v$

$v_s=(\frac{723}{697}-1)\cdot 343$

$v_s=12.79\approx 12.8\ m/s$

i.e.fork acquired a velocity of $12.8 m/s$

distance traveled by fork is given by

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

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s=displacement

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2 years ago

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

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Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

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the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

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$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

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Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

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2 years ago

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Answer:

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2 years ago