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3 years ago

7

Why is it hard to breathe when you go higher up

1 answer:

Fantom [35]3 years ago

8 0

When we breathe in air at sea level, the atmospheric pressure of about 14.7 pounds per square inch (1.04 kg. per cm.2) causes oxygen to easily pass through selectively permeable lung membranes into the blood. At high altitudes, the lower air pressure makes it more difficult for oxygen to enter our vascular systems.

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A polarized light that has an intensity I0 = 60.0 W/m² is incident on three polarizing disks whose planes are parallel and cente

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Answer:

The transmitted intensity through all polarizers is $I_3 =41.31 W/m^2$

Explanation:

According to Malu's law the intensity of a polarized light having an initial intensity $I_0$ is mathematically represented as

$I = I_0cos^2 \theta$

Now considering the polarizer(The polarizing disk) the equation above becomes

$I = I_0 (cos^2 \theta)^n$

Where n is the number of polarizers

Substituting $60.0W/m^2$ for the initial intensity 3 for the n and 20° for the angle of rotation

$I_3 = 60 (cos^220)^3$

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Which of the following is true about all electromagnetic waves in a vacuum

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The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the

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Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

<em>P </em>sun = <em>I </em><em>sun-earth </em><em>A </em><em>sun-earth</em>

where A = 2.863 10²³×m², and <em>I </em> is 1360 W/m²

<em>P </em>sun = 2.863 10²³ × 1360

<em>P </em>sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

<em>I </em><em>sun-mars = </em><em>P </em>sun / A sun-mars

where <em>P </em>sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>3.85×10²⁶W / 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>589.6 ≈ 590 W/m²

<em>I </em><em>sun-mars = </em>590 W/m²

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