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2 years ago

Sound waves are converted into mechanical movements by the

Physics

1 answer:

iren2701 [21]2 years ago

6 0

Answer:

a.auditory ossicles

b.oval window

c.Round window

d.tympanic membrane

Answer is tympanic membrane

Explanation:

The tympanic membrane otherwise called the ear drum is a membrane shaped like a cone,it connects the outside to the inner ear,it serves to convert vibration from air into fluid membrane vibration a good example of mechanical waves for onwatd transmission into the cochlea of the inner ear through the oval window

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A 615.00 kg race car is uniformly traveling around a circular race track. It takes the race car 20.00 seconds to do one lap arou

koban [17]

Answer:

The value is $f = 0.05 \ Hz$

Explanation:

From the question we are told that

The mass of the car is $m = 615 \ kg$

The period of the circular motion is $T = 20 \ s$

The radius is $r = 80 \ m/s$

Generally the frequency of the circular motion is

$f = \frac{1}{T }$

=> $f = \frac{1}{ 20 }$

=> $f = 0.05 \ Hz$

3 0

3 years ago

When it gets that hot the evergreens can explode?

Ghella [55]

If you mean the tree, evergreen trees can exploded if theres extreme stress on the trunk

3 0

3 years ago

. Write the following in the scientific notation of 0.0000002400m

aleksandr82 [10.1K]

Explanation:

<h2>Steps :</h2>

Move decimal from left to right =0 000000240.0
Then count the numbers before decimal and write it like this =240.0x10 power-9
That's all

hope it help

7 0

2 years ago

Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at

PSYCHO15rus [73]

Answer:

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

$\mu= \frac{m_1m_2}{m_1+m_2}$

now, since both the masses have mass m

therefore,

$\mu= \frac{m^2}{2m}$

= m/2

The final K.E. of the particles is

$KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2$

Distance between two particles is d and the gravitational potential energy between them is given by

$PE_{Gravitational}= \frac{Gmm}{d}$

By law of conservation of energy we have

$KE_{initial}+KE_{final}= PE_{gravitaional}$

Now plugging the values we get

$0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}$

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

$=\sqrt{\frac{Gm}{d} }$

This the required relation between G,m and d

5 0

2 years ago

What is the frequency of an x- ray if the wavelength is 4.5 E - 10m?

AnnZ [28]

V (speed) = F (frequency) x Wavelength
If we rearrange the formula, making frequency the subject;
F (frequency) = Speed ÷ Wavelength
F = 300,000 m\s x 4.5 e -10m
F = 0.08810409956 Hz

4 0

2 years ago