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abruzzese [7]
3 years ago
11

What is the frequency of sound waves whose wavelength is 0.25 m on a day when the air temperature is 25 degree Celsius? Group of

answer choices
1200 Hz
1384 Hz
1730 Hz
2400 Hz
3000 Hz
Physics
2 answers:
anzhelika [568]3 years ago
8 0

Answer:

Frequency, f = 1384 Hz                                                                

Explanation:

It is given that,

Wavelength of the sound wave, \lambda=0.25\ m

Air temperature, T = 25° C

The speed of sound at a particular temperature is calculated as :

v=331.5+0.6\times T

v=331.5+0.6\times 25

v = 346.5 m/s

Let f is the frequency of the sound wave. It can be calculated as :

f=\dfrac{v}{\lambda}

f=\dfrac{346.5\ m/s}{0.25\ m}

f = 1386 Hz

or

f = 1384 Hz

So, the frequency of sound waves is 1384 Hz. Hence, this is the required solution.

Dvinal [7]3 years ago
5 0

Answer:1384 Hz

Explanation:

Given

wavelength(\lambda )=0.25 m

Temperature T=25^{\circ}

at T=25^{\circ} velocity of sound is 346 m/s

and we know

velocity=frequency\times \lambda

v=f\times \lambda

346=f\times 0.25

f=1384 Hz

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A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th
Stolb23 [73]

Answer:

The current is  I_b  =  400 \ A

Explanation:

From the question we are told that

    The  length of the segment is  l  =  2.50  \  m

     The current is  I_a  =  1000 \ A

     The force felt is  F  =  4.0 \  N

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Generally the current on the second wire is mathematically represented as

        I_b  =  \frac{2 \pi * r * F }{ l *  \mu_o  *  I_a }

Here  \mu_o is the permeability of free space with value  \mu_o =  4 \pi * 10^{-7} \ N/A^2

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4 0
3 years ago
A tennis player hits a ball 2.0 m above the ground.
tangare [24]

Explanation:

initial height, yo = 2 m

initial velocity, u = 20 m/s

angle of projection,θ = 5 degree

distance of net = 7 m

height of net = 1 m

Let it covers a vertical distance y in time t .

Use Second equation of motion for vertical motion

As it hits the ground in time t, so put y = 0

Taking positive sign, t = 0.84 s

The ball travels a horizontal distance x in time t

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As this distance is more than the distance of net, so it clears the net.

Let t' be the time taken to travel a horizontal distance equal to the distance of net

7 = 20 cos5 x t'

t' = 0.35 s

Let the vertical distance traveled by the ball in time t' is y'.

So,

y' = 2.008 m

So, it clears the net which is 1 m high.

It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m

your welcome, and have a great day.

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