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otez555 [7]
2 years ago
15

Calculate the intrapleural pressure if atmospheric pressure is 765 millimeters of mercury, assuming that the subject is at rest

(not inhaling or exhaling).
A. 4 millimeters of mercury
B. 0 millimeters of mercury
C. 761 millimeters of mercury
D. 765 millimeters of mercury
Physics
1 answer:
dimaraw [331]2 years ago
5 0

Answer:

C. 761 millimeters of mercury

Explanation:

Because it is at rest, the intrapulmonary pressure must be equal to the atmospheric pressure. And the intrapleural pressure is usually 4 millimeters of mercury (mmHg) less than the intrapulmonary pressure. So we have the following

intrapulmonary pressure = 765 millimeters of mercury

intrapleural pressure = intrapulmonary pressure - 4 millimeters of mercury

Then:

intrapleural pressure = (765 -4) millimeters of mercury

intrapleural pressure = 761 millimeters of mercury

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VikaD [51]
Your answer is C.

hope this helps!
8 0
2 years ago
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Tarzan swings back and forth on a long vine with a period of 7.27 s. how long is the vine?(unit=m)
lana [24]

Answer:

Tarzan, who weighs 688N, swings from a cliff at the end of a convenient vine that is 18m long. From the top of the cliff to the bottom of the swing he descends by 3.2m.

Explanation:

3 0
2 years ago
the mass of piece of granite is 15.6g when it is suspended can a mass of 5.5g of water is displaced find the dencity of granite
Blababa [14]

Answer:

2836.36 kg/m³

Explanation:

Applying,

Densty of granite(D')/Density of water(D) = weight of granite(W')/weight of water displaced(W)

D'/D = W'/W................... Equation 1

Make D' the subject of the equation

D' = W'D/W............... Equation 2.

From the question,

Given: W' = mg = 9.8(15.6/1000) = 0.15288 N, W = 9.8(5.5/1000) = 0.0539 N, Constant: D = 1000 kg/m³

Substitute these values into equation 2

D' = 1000(0.15288)/0.0539

D' = 2836.36 kg/m³

3 0
1 year ago
Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is
GarryVolchara [31]

that would be given by

[email protected]

@ representing coefficient of kinetic friction.

thus 19.5/51.7 = 0.377

6 0
2 years ago
A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision
TEA [102]

Answer:

D) No, since kinetic energy is not conserved.

Explanation:

Since momentum is always conserved in all collision

so in Y direction we can say

0 = m(3.5 sin30) - mv_y

v_y = 1.75 m/s

Now similarly in X direction we will have

m(4.5) = m(3.5 cos30 ) + mv_x

v_x = 1.47 m/s

now final kinetic energy of both puck after collision is given as

KE_f = \frac{1}{2}m(3.5^2) + \frac{1}{2}m(1.75^2 + 1.47^2)

KE_f = 8.73 m

initial kinetic energy of both pucks is given as

KE_i = \frac{1}{2}m(4.5^2) + 0

KE_i = 10.125 m

since KE is decreased here so it must be inelastic collision

D) No, since kinetic energy is not conserved.

5 0
2 years ago
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