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Ronch [10]
3 years ago
6

Air that enters the pleural space during inspiration but is unable to exit during expiration creates a condition called a. open

pneumothorax. b. empyema. c. pleural effusion. d. tension pneumothorax.
Physics
1 answer:
sertanlavr [38]3 years ago
4 0

Answer:

The correct answer is d. tension pneumothorax.

Explanation:

The increasing build-up of air that is in the pleural space is what we call the tension pneumothorax and this happens due to the lung laceration that lets the air to flee inside the pleural space but it does not return.

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Electromagnets are created by?
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A coil of insulated wire around an iron core 
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How did the continental Shelf form? Please help, thank you!! :)
shutvik [7]
The movements of the tectonic plates
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3 years ago
Calculate the binding energy per nucleon of the helium nucleus 52he. express your answer in millions of electron volts to four s
Vitek1552 [10]
The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for <span>52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=</span><span>binding energy
</span><span>52He-----------> 2 x 11p + 3 x10n is the equation bilan
</span>so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
                                                        mn=mass of neutron=<span>1.67 10^-27 kg
                                                        </span><span>m nucleus= 5
Dm= 2x</span>1.67 10^-27 kg+ 3x<span>1.67 10^-27 kg-5=  - 4.9 J
Eb= </span> - <span>4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
so the answer is Eb /nucleon = Eb/5= -9.10^16 J, but 1eV=1.6 . 10^-19 J
so </span><span>-9.10^16 J/ 1.6 10^-19=  -5.625 10^35 eV

the final answer is </span><span>Eb /nucleon </span><span>=  -5.625 x10^35 eV</span>
8 0
3 years ago
A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the o
katen-ka-za [31]

Answer:

<h2>42.67N</h2>

Explanation:

Step one:

<u>Given </u>

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

<u>Required</u>

Force F

the expression for the force is

F=mΔv/t

F=0.32*(22-14)/0.06

F=(0.32*8)/0.06

F=2.56/0.06

F=42.67N

The average force exerted on the bat 42.67N

4 0
3 years ago
At a distance 30 m from a jet engine, intensity of sound is 10 W/m^2. What is the intensity at a distance 180 m?
AleksandrR [38]

Answer:

I_{2}=0.27 W/m^2

Explanation:

Intensity is given by the expresion:

I_{2}=Io (\frac{r1}{r2} )^{2}

where:

Io = inicial intensity

r1= initial distance

r= final distance

I_{2}=10 W/m^2 (\frac{30m}{180m} )^{2}

I_{2}=0.27 W/m^2

5 0
3 years ago
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