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bonufazy [111]
3 years ago
11

Two identical loudspeakers that are 5.00 m apart and face toward each other are driven in phase by the same oscillator at a freq

uency of 875 Hz. The speed of sound in the room is 344 m/s. If you start out standing midway between the speakers, find the shortest distance you can walk toward either speaker in order to hear a minimum of sound.
Physics
1 answer:
Lunna [17]3 years ago
8 0

Answer:0.0982 m

Explanation:

Given

distance between two loudspeaker is 5 m

frequency (f )=875 Hz

speed of sound (v)=344 m/s

Let x_0 be the smallest distance moved by observer then

Position of observer w.r.t to first speaker is

x_1=\frac{L}{2}-x_0

Position of observer w.r.t to second speaker is

x_2=\frac{L}{2}+x_0

\Delta x=2x_0

For Destructive interference

\Delta x=\left (  m+\frac{1}{2}\right )\cdot lambda

For minimum m=0

and \lambda =\frac{v}{f}

2x_0=\frac{v}{2f}

x_0=\frac{v}{4f}=\frac{344}{4\times 875}=0.0983 m

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3 years ago
When an object is falling and reaches a constant velocity, the net force on the object is ____ and the weight of the object is e
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When an object is falling and reaches a constant velocity, the net force on the object is <em>zero</em> (it's not accelerating), and the weight of the object is equal to <em>the force of air resistance against the object</em>.  (choice-D)

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A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa
andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

\omega=\frac{v}{r}

where

\omega is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

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(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

\omega=\frac{v}{r}

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s

(c) 5550 m

The maximum playing time of the CD is

t =74.0 min \cdot 60 s/min = 4,440 s

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

d=vt=(1.25 m/s)(4,440 s)=5,550 m

(d) -6.4\cdot 10^{-3} rad/s^2

The angular acceleration of the CD is given by

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f = 21.6 rad/s is the final angular speed (when the CD is scanned at the outermost part)

\omega_i = 50.0 rad/s is the initial angular speed (when the CD is scanned at the innermost part)

t=4440 s is the time elapsed

Substituting into the equation, we find

\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2

5 0
3 years ago
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A yard is shorter than a meter.

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>>>1 mile is 1.609 km, so a mile is longer than a kilometer

>>>1 foot is 30.48cm, so a foot longer than a centimeter

>>> 1 inch is 2.54cm, so an inch is longer than a centimeter

From the above relationships, only a yard is shorter than a meter is true. Others are wrong.

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It is the study of cognitive functions.
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