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2 years ago

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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A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a sp

dusya [7]

from the question you can see that some detail is missing, using search engines i was able to get a similar question on "https://www.slader.com/discussion/question/a-student-throws-a-water-balloon-vertically-downward-from-the-top-of-a-building-the-balloon-leaves-t/"

here is the question : A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?

Answer:

(A) 26 m/s

(B) 32.4 m

(C) v = 15.4 m/s

Explanation:

initial speed (u) = 6.4 m/s

acceleration due to gravity (a) = 9.9 m/s^[2}

time (t) = 2 s

(A) What is its speed after falling for 2.00s?

from the equation of motion v = u + at we can get the speed

v = 6.4 + (9.8 x 2) = 26 m/s

(B) How far does it fall in 2.00s?

from the equation of motion $s=ut+0.5at^{2}$ we can get the distance covered

s = (6.4 x 2) + (0.5 x 9.8 x 2 x 2)

s = 12.8 + 19.6 = 32.4 m

c) What is the magnitude of its velocity after falling 10.0m?

from the equation of motion below we can get the velocity

$v^{2} = u^{2} + 2as\\v^{2} = 6.4^{2} + (2x9.8x10)\\V^{2} = 236.96\\v = \sqrt{236.96}$

v = 15.4 m/s

7 0

3 years ago

Barry is conducting an experiment and rolls a tennis ball down a ramp. Which best describes the motion of the tennis ball? It do

yawa3891 [41]

Answer:

A. It does not exhibit projectile motion and follows a straight path down the ramp.

7 0

2 years ago

Read 2 more answers

Which law states that the pressure and absolute temperature of a fixed quantity of gas are directly proportional under constant

Brrunno [24]

<u>Gay Lussac’s law</u> state that the pressure and absolute temperature of a fixed quantity of a gas are directly proportional under constant volume conditions.

<h2>Further Explanation </h2><h3>Gay-Lussac’s law </h3>

It states that at constant volume, the pressure of an ideal gas I directly proportional to its absolute temperature.
Thus, an increase in pressure of an ideal gas at constant volume will result to an increase in the absolute temperature.

<h3>Boyles’s law </h3>

This gas law states that the volume of a fixed mass of a gas is inversely proportional to its pressure at constant absolute temperature.
Therefore, when the volume of an ideal gas is increased at constant temperature then the pressure of the gas will also increase.

<h3>Charles’s law </h3>

It states that the volume of a fixed mass of a gas is directly proportional to absolute temperature at constant pressure.
Therefore, an increase in volume of an ideal gas causes a corresponding increase in its absolute temperature and vice versa while the pressure is held constant.

<h3>Dalton’s law </h3>

It is also known as the Dalton’s law of partial pressure. It states that the total pressure of a mixture of gases is always equivalent to the total sum of the partial pressures of individual component gases.
Partial pressure refers to the pressure of an individual gas if it occupies the same volume as the mixture of gases.

Keywords: Gas law, Gay-Lussac’s law, pressure, volume, absolute temperature, ideal gas

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