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grin007 [14]
3 years ago
5

How do you sound waves reach our ears?

Physics
1 answer:
iren2701 [21]3 years ago
5 0

Answer:

Sound waves enter the outer ear and travel through a narrow passageway called the ear canal, which leads to the eardrum. The eardrum vibrates from the incoming sound waves and sends these vibrations to three tiny bones in the middle ear. These bones are called the malleus, incus, and stapes.

Explanation:

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The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which
emmainna [20.7K]

Answer:

Explanation:

A ) angular velocity ω = 2π / T

= 2 x 3.14 / 60

= .10467 rad / s

linear velocity v = ω R

=  .10467 x 50

= 5.23 m / s

centripetal force = m v² / R

= mg v² / gR

= 834 x 5.23² / 9.8 x 50

= 46.55 N

B )

apparent weight

= mg - centripetal force

= 834 - 46.55

= 787.45 N

C ) apparent weight

= mg + centripetal force

= 834 + 46.55

= 880.55 N.

D )

For apparent weight to be zero

centripetal force = mg

mg = mv² / R

v² = gR

= 9.8 x 50

= 490

v = 22.13 m /s

time period of revolution

= 2π R /v

2 x 3.14 x 50 / 22.13

= 14.19 s  

8 0
3 years ago
To understand the formula for power radiated in the form of electromagnetic energy by an object at nonzero temperature. every ob
lbvjy [14]

As per Stefan - Boltzmann law we know that

1. Power radiated in the form of electromagnetic energy by an object at nonzero temperature.

2. Every object at absolute (kelvin) temperature t will radiate electromagnetic waves.

3. This radiation is typically in the infrared for objects at room temperature, with some visible light emitted for objects heated above 1000 k.

4. The formula governing the rate of energy radiation from a surface is given by p=eσat^4,

where p is the thermal power (also known as the heat current h).

Thermal radiation in visible light can be seen on hot metalwork. Its emission in the infrared is invisible to the human eye. Infrared cameras are capable of capturing this infrared emission.

Thermal radiation is electromagnetic radiation generated by the thermal motion of charged particles in matter. All matter with a temperature greater than absolute zero emits thermal radiation. Particle motion results in charge-acceleration or dipole oscillation which produce electromagnetic radiation.

Examples of thermal radiation include the visible light and infrared light emitted by an incandescent light bulb, the infrared radiation emitted by animals that is detectable with an infrared camera, and the cosmic microwave background radiation. Thermal radiation is different from thermal convection and thermal conduction—a person near a raging bonfire feels radiant heating from the fire, even if the surrounding air is very cold.

Sunlight is part of thermal radiation generated by the hot plasma of the Sun. The Earth also emits thermal radiation, but at a much lower intensity and different spectral distribution. The Earth's absorption of solar radiation, followed by its outgoing thermal radiation, are the two most important processes that determine the temperature and climate of the Earth in most climate models.

So the correct answer which is applicable here will be

This formula applies to any object of total surface area a, kelvin temperature t, and emissivity e

here

\sigma[\tex] = stefan boltzmann constant = [tex]5.67 * 10^{-8}

3 0
3 years ago
The electric potential at the dot in the figure is 3160 V. What is charge q?
PSYCHO15rus [73]

Hi there!

Recall the equation for electric potential of a point charge:

V = \frac{kQ}{r}

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V

Lower left charge's potential:

V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V

Add the two, and subtract from the total EP at the point:

3160 + 1123.75 - 2247.5 = 2036.25


The remaining charge must have a potential of 2036.25 V, so:

2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}


5 0
2 years ago
A race car traveling at 10 meters per second accelerates at 1.5 meters per second squared while moving a distance of 600 meters.
Mekhanik [1.2K]

Answer:

Explanation:

Givens

vi = 10 m/s

a = 1.5 m/s^2

d = 600 m

vf = ?

Formula

vf^2 = vi^2 + 2*a*d

Solution

vf^2 = 10^2 + 2*1.5 * 600

vf^2 = 100 + 1800

vf^2 = 1900

sqrt(vf^2) = sqrt(1900)

vf = 43.59 m/s

7 0
3 years ago
A spectral line from a star is observed to have a wavelength of 656.5 nm. The rest wavelength of this line is 656.8 nm. (a) What
nignag [31]

Answer:

speed of star is  1.37 × 10^{5} m/s

it is approaching to earth because wavelength of star is decreasing than rest

if emit same wavelength it does not move anywhere

it will remain steady condition with respect earth

Explanation:

given data

wavelength = 656.5 nm

rest wavelength = 656.8 nm

to find out

the speed of the star , is it approaching

solution

we know here equation that is

wavelength shift / wavelength at rest  = velocity of source / speed of light

so put all value and find

velocity of source = 3 × 10^{-9} × 3 × 10^{8} / 656.8 × 10^{-9}

velocity of source star = 1.37 × 10^{5} m/s

and

it is approaching to earth because wavelength of star is decreasing than rest

and

if emit same wavelength it does not move anywhere

it will remain steady condition with respect earth

6 0
3 years ago
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