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3 years ago

How does an airbag help protect a passenger in a car accident?

2 answers:

8 0

Good question. It like a fail safe,if you hit a car tail gate going slightly fast it come out to protect you from hitting your on the the wheel or the mirror. See if you don't have a seat belt or it just don't fully stop you the airbag might help.Like concussions. I Hope This Help you :)

likoan [24]3 years ago

3 0

Answer:

It shortens the passengers stopping time, and this decreases the force.

I think.

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Answer:

The orbital speed can be found using v = SQRT(G*M/R). The R value (radius of orbit) is the earth's radius plus the height above the earth - in this case, 6.59 x 106 m.

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3 years ago

One day in second grade, Chris drank chocolate milk before recess. She then played hard during recess and got sick from the heat

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Answer:

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3 years ago

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T or F-Climate is directly related to the amount of energy from the sun or solar energy that an area receives

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Answer:

true

Explanation:

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3 years ago

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Dmitry [639]

equal and opposite reaction.

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3 years ago

A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr

rjkz [21]

Answer:

Part A

I = 1.4 mW/m²

Part B

β = 91.46 dB

Explanation:

Part A

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^{2}}$

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

$I = \frac{34}{4\pi (44)^{2}}$

I = 0.0013975 W/m²

≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

Part B

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;

β $= 10log_{10}\frac{I}{I_{0}}$ dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

∴ β $= 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}$

β $= 10log_{10} (1.4 * 10^{9})$

β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.

4 0

3 years ago