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sashaice [31]
3 years ago
8

Which of the following pressures rises and falls with the phases of breathing but eventually equalizes with the pressure of the

air in the environment?
A) atmospheric pressure
B) intrapleural pressure
C) intrapulmonary pressure
D) transpulmonary pressur
Physics
1 answer:
VikaD [51]3 years ago
7 0

Answer:

(c) Intrapulmonary pressure

Explanation:

Intrapulmonary pressure is the pressure of air within the alveoli, it changes with the different phases of breathing, and because it is connected to the atmosphere through the throat if eventually equalizes with the atmospheric air pressure in the environment.

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7. You start to walk toward the east towards home at a constant speed of 4 km/hr. At the same time, Someone else leaves your hom
amm1812

A) position time graph for both is shown

here one of the graph is of lesser slope which means it is moving with less speed while other have larger slope which shows larger speed

At one point they intersects which is the point where they both will meet

B) Let the two will meet after time "t"

now we can say that

if they both will meet after time "t"

then the total distance moved by you and other person will be same as the distance between you and home

so it is given as

v_1t + v_2t = d

4*t + 28*t = 3.2 km

t = \frac{3.2}{32} = 0.1 hr

so they will meet after t = 6 min

so from position time graph we can see that two will meet after t = 6 min where at this position two graphs will intersect


4 0
4 years ago
Hiii please help i’ll give brainliest if you give a correct answer please thanks!
MaRussiya [10]

Answer:

i think number 4 :/ i hope its right...

6 0
3 years ago
A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a
Makovka662 [10]

Answer:

0.51 m

Explanation:

Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.

Kinetic energy, KE=½kx²

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Potential energy, PE=mgh

Where g is acceleration due to gravity, h is height and m is mass

Equating KE=PE

mgh=½kx²

Making x the subject of formula

x=\sqrt {\frac {2mgh}{k}}

Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then

x=\sqrt \frac {2*1300*9.81*10}{1000000}=0.50503465227646m\\x\approx 0.51 m

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3 years ago
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8 0
4 years ago
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It is actually are tools that make work easier C because i just had it on study island

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