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Alexeev081 [22]

3 years ago

14

an object is producing a sound that has a wavelength in air of 2.69m. If the speed of sound in air is 346m/s, what is the freque

ncy of the sound produced by the object ?

1 answer:

Misha Larkins [42]3 years ago

5 0

Answer:

129.74 Hz

Explanation:

Given:

Wave velocity ( v ) = 346 m / sec

wavelength ( λ ) = 2.69 m

We have to calculate Frequency ( f ) :

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > f = v / λ

Putting values here we get:

= > f = 346 / 2.69 Hz

= > f = 34600 / 269 Hz

= > f = 129.74 Hz

Hence, frequency of sound is 129.74 Hz.

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Answer:

Tarzan, who weighs 688N, swings from a cliff at the end of a convenient vine that is 18m long. From the top of the cliff to the bottom of the swing he descends by 3.2m.

Explanation:

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A police siren of frequency fsiren is attached to a vibrating platform. The platform and siren oscillate up and down in simple h

babunello [35]

Answer:

he maximum frequency occurs when the denominator is minimum

f’= f₀ $\frac{343}{343 + v_s}$

Explanation:

This is a doppler effect exercise, where the sound source is moving

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3 years ago

A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming

Monica [59]

Answer:

The deceleration is $a = - 76.27 m/s^2$

Explanation:

From the question we are told that

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The length by which the net is stretched is $s = 1.8 \ m$

From the law of energy conservation

$KE_T + PE_T = KE_B + PE_B$

Where $KE_T$ is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

and $PE_T$ is the potential energy of the before jumping which is mathematically represented at

$PE_T = mg H$

and $KE_B$ is the kinetic energy of the person just before landing on the safety net which is mathematically represented at

$KE_B = \frac{1}{2} m v^2$

and $PE_B$ is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

So the above equation becomes

$mgH = \frac{1}{2} m v^2$

=> $v = \sqrt{2 gH }$

substituting values

$v = 16.57 m/s$

Applying the equation o motion

$v_f = v + 2 a s$

Now the final velocity is zero because the person comes to rest

So

$0 = 16.57 + 2 * a * 1.8$

$a = - \frac{16.57^2 }{2 * 1.8}$

$a = - 76.27 m/s^2$

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3 years ago

A force of 10 N causes a spring to extend by 20 mm. Find: a) the spring constant of the spring in N/m

Mars2501 [29]

Answer:

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2 years ago

A body is dropped from the roof of a 20 m high building by how much:

USPshnik [31]

Answer:

t = 2.01 s

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Explanation:

It's know through the International System that the earth's gravity is 9.8 m/s², then we have;

Data:

Height (h) = 20 m
Gravity (g) = 9.8 m/s²
Time (t) = ?
Final Velocity (Vf) = ?

==================================================================

Time

Use formula:

$\boxed{t=\sqrt{\frac{2*h}{g}}}$

Replace:

$\boxed{t=\sqrt{\frac{2*20m}{9.8\frac{m}{s^{2}}}}}$

Everything inside the root is solved first. So, we solve the multiplication of the numerator:

$\boxed{t=\sqrt{\frac{40m}{9.8\frac{m}{s^{2}}}}}$

It divides:

$\boxed{t=\sqrt{4.08s}}$

The square root is performed:

$\boxed{t=2.01s}$

==================================================================

Final Velocity

use formula:

Vf = g * t

Replace:

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Multiply:

Vf = 19.7 m/s

==================================================================

How long does it take to reach the ground?

Takes time to reach the ground in <u>2.01 seconds.</u>

How fast does it hit the ground?

Hits the ground with a speed of <u>19.7 meters per seconds.</u>

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3 years ago