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2 years ago

Why is it less air pressure up in the sky and why does it affect our ears

Physics

1 answer:

IgorLugansk [536]2 years ago

8 0

Answer:

When you in a airplane you are in high altitude and thats where air density drops. Thats where ear pops come from.

Explanation:

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The loudness l of a sound, measured in decibels, is given by l=10log10r, where r is the sound's relative intensity. suppose one

marishachu [46]

Answer

given,

I is the loudness of sound

I = 10 Log₁₀ r

r is relative intensity

at when relative intensity is 10⁶

I = 60 dB

how much louder when 100 people would be talking together

I = 10 Log₁₀ r

I = 10 Log₁₀ (10⁶ x 100)

I = 10 Log₁₀ (10⁸)

I = 80 dB

hence, the intensity will be increased by (80 dB -60 dB) 20 dB when 100 people start talking together.

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3 years ago

ILL GIVE BRAINLYEST

lawyer [7]

Answer:

3rd picture straight line going up right

Explanation:

3rd picture

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3 years ago

Suppose a straight 1.00-mm-diameter copper wire could just "float" horizontally in air because of the force due to the Earth’s m

insens350 [35]

To solve this problem it is necessary to apply the concepts related to the Force since Newton's second law, as well as the concept of Electromagnetic Force. The relationship of the two equations will allow us to find the magnetic field through the geometric relations of density and volume.

$F_{mag}= BIL$

Where,

B = Magnetic Field

I = Current

L = Length

Note: $F_{mag}$ is a direct adaptation of the vector relation $F=q \times V \times B$

From Newton's second law we know that the relation of Strength and weight is determined as

$F_g = mg$

Where,

m = Mass

g = Gravitational Acceleration

For there to be balance the two forces must be equal therefore

$F_{mag} = F_g$

$BIL = mg$

Our values are given as,

Diameter $(d) = 1.0mm = 1*10^{-3}m$

Radius $(r) = \frac{d}{2} = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m$

Magnetic Field $(B) = 5.0*10^{-5} T$

From the relationship of density another way of expressing mass would be

$\rho = \frac{m}{V} \rightarrow m = \rho V$

At the same time the volume ratio for a cylinder (the shape of the wire) would be

$V = \pi r^2 L \rightarrow L =Length, r= Radius$

Replacing this two expression at our first equation we have that:

$BIL = mg$

$BIL = ( \rho V)g$

$BIL = ( \rho \pi r^2 L)g$

Re-arrange to find I

$I = \frac{( \rho \pi r^2 L)g}{BL}$

$I = \frac{( \rho \pi r^2 )g}{B}$

We have for definition that the Density of copper is $8.9*10^3 Kg/m^3$ , gravity acceleration is $9.8m/s^2$ and the values of magnetic field (B) and the radius were previously given, then: