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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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The atoms in a solid move about freely
ivolga24 [154]

No, not exactly.  They jiggle and tremble and vibrate a lot, but
they always basically stay in very nearly the same place.

It's like if you're allowed to go anywhere you want in your jail cell,
you wouldn't exactly call that "moving about freely".

6 0
3 years ago
Consider an object rolling off of your lab bench. Discuss how you might be able to make measure- ments to determine its initial
guapka [62]

Answer:

measure the position every so often with a stopwatch

Explanation:

A possible method of measurement is to place a measuring tape along the path and measure the position every so often with a stopwatch, with this we can make a graph of position against time and by extrapolation find the initial velocity.

This is a method used in measurements of uniform movements of bodies

7 0
3 years ago
Which ratio is greater than 5/8 <br><br> A)12/24 <br><br> B)3/4 <br><br> C)15/24 <br><br> D)4/12
pav-90 [236]

12/24=4x3/8x3

=4/8

4/8 is not correct ans it is less than 5/8

B) 3/4

multiply and divide it by 2 so the denominator should be equal to 8

3x2/4x2=6/8

6/8 is greater that 5/8 hence b is correct option

C)15/24=5/8 which is equalt to 5/8 hence it is not correct answer

D)4/12=1/3 it is also less than 5/8


5 0
3 years ago
Read 2 more answers
A solid cylinder of radius 10.0 cm rolls down an incline with slipping. The angle of the incline is 30°. The coefficient of kine
erik [133]

To solve this problem it is necessary to apply the expressions related to the calculation of angular acceleration in cylinders as well as the calculation of linear acceleration in these bodies.

By definition we know that the angular acceleration in a cylinder is given by

\alpha = \frac{2\mu_k g cos\theta}{r}

Where,

\mu_k = Coefficient of kinetic friction

g = Gravitational acceleration

r= Radius

\theta= Angle of inclination

While the tangential or linear acceleration is given by,

a = g(sin\theta-\mu_k cos\theta)

ANGULAR ACCELERATION, replacing the values that we have

\alpha = \frac{2\mu_k g cos\theta}{r}

\alpha = \frac{2(0.4)(9.8) cos(30)}{10*10^{-2}}

\alpha = 67.9rad/s

LINEAR ACCELERATION, replacing the values that we have,

a = (9.8)(sin30-(0.4)cos(30))

a = 1.5m/s^2

Therefore the linear acceleration of the solid cylinder is 1.5 m/s^2

8 0
3 years ago
All of the following help to reduce the problem of social traps except ___________. Please select the best answer from the choic
allsm [11]
It’s c beside me in the
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