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uysha [10]
3 years ago
7

What happens if the air pressure in your throat and outside body is less than the air pressure in your middle ear when you swall

ow
Physics
1 answer:
andrezito [222]3 years ago
8 0

Answer:

When the air pressure in the throat and outside the body is less than the air pressure in the middle ear, barotrauma occurs.

Explanation:

Ear barotrauma is a medical condition that describes discomfort in the ear which is caused by pressure differences in the inner and outer ear drum.

Usually, the air pressure in the middle ear is the same as the air pressure in the throat and outside the body.

When we swallow, the eustachian tube opens up and air flows out of and into the middle ear, this balances the pressure. But if the eustachian tube is blocked, the air pressure in the throat and outer body become different from the air pressure in the middle ear.

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Visible light or electromagnetic radiation within 400nm to 700nm is responsible for colour of the spectrum.

Explanation:

The electromagnetic spectrum contains radiations of varying wavelength. The radiations with the lowest energy are characterised by the longest wavelength.

Within this spectrum lies the visible light which enables us to see a different colour. The radiations within the range 400nm to 700nm are included in the visible spectrum.

While violet lies at the 400nm spectrum part red colour lies at 700nm part. As the wavelength of the radiation transverses between 400-700 nm, the colour of the object changes accordingly.

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What is the key characteristic for double displacement
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A 500 μF capacitor is wired in series with a 5 V battery and a 20 kΩ resistor. What is the voltage across the capacitor after 20
liq [111]

Answer:

Explanation:

It is a question relating to charging of capacitor . For charging of capacitor , the formula is as follows.

Q = CV ( 1 - e^{-\lambda\times t} )

λ = 1/CR , C is capacitance and R is resistance.

= 1/(500 x 10⁻⁶ x 20 x 10³ )

= .1

λ t = .1 x 20

λ t = 2

CV = 500 X 10⁻⁶ X 5

= 2500 X 10⁻⁶ C

Q = 2500 x 10⁻⁶ ( 1 - e^{-2} )

= 2500 x 10⁻⁶ x .86566

= 2161.66 μ C .

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A 3.00-m rod is pivoted about its left end. A force of 7.80 N is applied perpendicular to the rod at a distance of 1.60 m from t
dalvyx [7]

Answer:

The net torque about the pivot is and the answer is 'c'

c. T_{net}=8.58

Explanation:

T=F*d

The torque is the force apply in a distance so it is the moment so depends on the way to be put it the signs so:

T_1=F_1*d_1

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T_{net}=12.48N-3.9N=8.58N

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