Ask question

Ask question

All categories

2 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 ...?

1 answer:

Feliz [49]2 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!
</span>

You might be interested in

The first stage in the GAS model of stress is

Vladimir [108]

<span>The first stage in the Gas model of stress is alarm and mobilization. So the correct option in regards to the given question is option “d”. Hans Selye is the person that evolved this model and he has explained this model in complete details. He has broken down his model into three stages. The first stage involves alarm and mobilization. The second stage includes resistance. The third and the final stage include the exhaustion stage. These are the stages that an organism goes through to restore back the balance when stress is exerted from outside. </span>

8 0

3 years ago

If the accepted value of a wave is 121 m/s, who has the most accurate method of measuring the speed of a wave?

qaws [65]

The answer would be erin out of all of them thank me later :)

5 0

2 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

A yellow and green car traveled 400 miles to Dayton, OH. The green car made the trip in 10 hours. The yellow arrived in 8 hours.

elixir [45]

Answer: C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

Explanation:

The speed of each car is defined as:

$v=\frac{d}{t}$

where d is the distance traveled by the car and t is the time taken.

For the yellow car, d=400 mi and t=8 h, so its speed is

$v=\frac{400 mi}{8 h}=50 mph$

For the green car, d=400 mi and t=10 h, so its speed is

$v=\frac{400 mi}{10 h}=40 mph$

So, the correct choice is

C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

4 0

3 years ago

Read 2 more answers

Assume the equation x 5 At3 1 Bt describes the motion of a particular object, with x having the dimension of length and t having

igomit [66]

Answer:

(a) A = m/s^3, B = m/s.

(b) dx/dt = m/s.

Explanation:

(a)

$x = At^3 + Bt\\m = As^3 + Bs\\m = (\frac{m}{s^3})s^3 + (\frac{m}{s})s$

Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.

(b) $\frac{dx}{dt} = 3At^2 + B = 3(\frac{m}{s^3})s^2 + \frac{m}{s} = m/s$

This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.

6 0

2 years ago