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RideAnS [48]
3 years ago
12

Which of the following creates an adhesive force that prevents separation of the parietal and visceral pleurae during ventilatio

n?
negative intrapulmonary pressure
negative intrapleural pressure
alveolar fluid surface tension
lung elasticity
Physics
1 answer:
Diano4ka-milaya [45]3 years ago
3 0

Answer:

Negative intrapleural pressure is the correct answer

Explanation:

Intrapleural pressure is more subatmospheric in the uppermost part of the thorax than in the lowermost parts in the standing horse.

Air moves from a region of higher pressure to one of lower pressure.  Therefore, for air to be moved into or out of the lungs, a pressure difference between the atmosphere and the alveoli must be established. If there is no pressure difference, no airflow will occur.

Under normal circumstances, inspiration is accomplished by causing alveolar pressure to fall below atmospheric pressure. When the mechanics of breathing are being discussed, atmospheric pressure is conventionally referred to as 0 cm H2O, so lowering alveolar pressure below atmospheric pressure is known as negative-pressure breathing.

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A kite is 100m above the ground. if there are 200m of string out what is the angle between the string and the horizontal? (assum
jeka57 [31]

Answer :  The angle between the string and the horizontal is 30 degrees

Explanation:  Imagine this a a triangle where the length of the string (200m) is the hypotenuse and the height of the kite is the opposite side (100m) .

Let the angle between the string and the horizontal be theta.

Now  sin (Theta) = opposite side/hypotenuse

                             =  100/200 = 1/2

Therefore Theta = Sin ⁻¹ ( 1/2 )

Theta = 30 degrees

4 0
4 years ago
What is a difference between a law and a hypothesis?
AnnZ [28]

The correct answer is

C ). A hypothesis includes an explanation for why two variables affect each other, but a law only describes how they affect each other.

8 0
3 years ago
Which of the following would have the most momentum?
Anarel [89]
I think it should be D as momentum is the product of mass and velocity...
4 0
3 years ago
A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the
Maurinko [17]

Answer:

(a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity R_{0}=10\ mCi

Time t_{1}=4\ hours

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

R=R_{0}e^{-\lambda t}

\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})

Put the value into the formula

\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})

\lambda=0.0000154\ s^{-1}

\lambda=1.55\times10^{-5}\ s^{-1}

We need to calculate the half life

Using formula of half life

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}

Put the value into the formula

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}

T_{\dfrac{1}{2}}=44.719\times10^{3}\ s

T_{\dfrac{1}{2}}=11.3\ hr

(b). We need to calculate the value of N₀

Using formula of N_{0}

N_{0}=\dfrac{3.70\times10^{6}}{\lambda}

Put the value into the formula

N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}

N_{0}=2.38\times10^{11}\ nuclei

(c). We need to calculate the sample's activity

Using formula of activity

R=R_{0}e^{-\lambda\times t}

Put the value intyo the formula

R=10e^{-(1.55\times10^{-5}\times30\times3600)}

R=1.87\ mCi

Hence, (a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

4 0
3 years ago
What is free nitrogen
Alenkasestr [34]

Answer:

It is simply molecular nitrogen (N2). Nitrogen, in its molecular form, consists of two nitrogen atoms bound together with a tripple bond

Explanation:

4 0
3 years ago
Read 2 more answers
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