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Anit [1.1K]
2 years ago
5

How does the diaphragm remove air from the lungs?

Physics
1 answer:
LekaFEV [45]2 years ago
4 0

Answer:

d) When the diaphragm pushes upward, it decreases the volume of the lungs, pushing air out.

Explanation:

the relaxed diaphragm muscle is dome-shaped, the contracted muscle is horizontal.

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Biofuel is fuel that is produced through contemporary processes from biomass, rather than by the very slow geological processes involved in the formation of fossil fuels, such as oil. Since biomass technically can be used as a fuel directly, some people use the terms biomass and biofuel interchangeably. 

Biofuel, any fuel that is derived from biomass—that is, plant or algae material or animal waste. Since such feedstock material can be replenished readily, biofuel is considered to be a source of renewable energy, unlike fossil fuels such as petroleum, coal, and natural gas.

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The awnser is B

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____, one of Saturn's icy moons, is unusual in the solar system in that it has volcanic activity that ejects plumes of icy parti
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Answer: Enceladus

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3 years ago
a liquid reactant is pumped through a horizontal, cylindrical, catalytic bed. The catalyst particles are spherical, 2mm in diame
natulia [17]

Answer:

The upper limit on the flow rate = 39.46 ft³/hr

Explanation:

Using Ergun Equation to calculate the pressure drop across packed bed;

we have:

\frac{\delta P}{L}= \frac{150 \mu_oU(1- \epsilon )^2}{d^2p \epsilon^3} + \frac{1.75 \rho U^2(1-\epsilon)}{dp \epsilon^3}

where;

L = length of the bed

\mu = viscosity

U = superficial velocity

\epsilon = void fraction

dp = equivalent spherical diameter of bed material (m)

\rho = liquid density (kg/m³)

However, since U ∝ Q and all parameters are constant ; we can write our equation to be :

ΔP = AQ + BQ²

where;

ΔP = pressure drop

Q = flow rate

Given that:

9.6 = A12 + B12²

Then

12A + 144B = 9.6       --------------   equation (1)

24A + 576B = 24.1    ---------------  equation (2)

Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So

288 B = 4.9

       B = 0.017014

From equation (1)

12A + 144B  = 9.6

12A + 144(0.017014) = 9.6

12 A = 9.6 - 144(0.017014)

A = \frac{9.6 -144(0.017014}{12}

A = 0.5958

Thus;

ΔP = AQ + BQ²

Given that ΔP = 50 psi

Then

50 = 0.5958 Q + 0.017014 Q²

Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;

Q² + 35.02Q - 2938.8 = 0

Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;

Q = 39.46 ft³/hr

3 0
3 years ago
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