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cricket20 [7]
3 years ago
5

Audio frequency range of a human ear is 20Hz - 20000 Hz. Express the range in terms of time period ? Answer needed urgently. Pl.

help. Thanks ! S.Ramya ...?
Physics
1 answer:
Feliz [49]3 years ago
5 0
Using the term c in this case is a little confusing. It is more generic to use a general velocity, v. That way, in this case, we know to use the speed of sound. 

wavelength*frequency=v 

wavelength_20Hz = (345 m/s)/(1/20s) 

<span>wavelength_20kHz = (345 m/s)/(1/20000s)


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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How does the size of a planet affected the amount and type of gas in its atmosphere
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3 0
3 years ago
Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu
11111nata11111 [884]

Answer:

velocity = 62.89 m/s  in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $ 10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $ 10^5$  kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $ 10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $ 10^5$

So, $ V_x = \frac{10^5}{3000} $  = $ \frac{100}{3} $  m/s

and $ V_y=\frac{160}{3}$  m/s

Therefore, velocity is = $ \sqrt{V_x^2 + V_y^2} $

                                   = $ \sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2} $

                                   = 62.89 m/s

And direction is

tan θ = $ \frac{V_y}{V_x}$     = 1.6

therefore, $ \theta = \tan^{-1}1.6 $

                   = $ 58 ^{\circ}$  from x-axis

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3 years ago
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