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.

Answers

Answer 1

Answer:

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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1. A sphere with a mass of 10 kg and radius of 0.5 m moves in free fall at sea level (where the air density is 1.22 kg/m3). If the object has a drag coefficient of 0.8, what is the object’s terminal velocity? What is the terminal velocity at an altitude of 5,000 m, where the air density is 0.736 kg/m3? Show all calculations in your answer.

Answers

The value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.

Given data:

The mass of sphere is, m = 10 kg.

The radius of sphere is, r = 0.5 m.

The density of air is, $\rho = 1.22 \;\rm kg/m^(3)$ .

The drag coefficient of object is, $C_(d)=0.8$ .

The altitude is, h = 5000 m.

The density of air at altitude is, $\rho' =0.736 \;\rm kg/m^(3)$ .

The mathematical expression for the terminal velocity of an object is,

$v_(t)=\sqrt(2mg)/(\rho * A * C_(d))$

here,

g is the gravitational acceleration.

A is the area of sphere.

Solving as,

$v_(t)=\sqrt{(2 * 10 * 9.8)/(1.22 * (4 \pi r^(2)) * C_(d))}\n\n\nv_(t)=\sqrt{(2 * 10 * 9.8)/(1.22 * (4 \pi * 0.5^(2)) * 0.8)}\n\n\n\v_(t)=7.99 \;\rm m/s$

Now, the terminal velocity at the altitude of 5000 m is given as,

$v_(t)'=\sqrt(2mg)/(\rho' * A * C_(d))$

Solving as,

$v_(t)'=\sqrt{(2 * 10 * 9.8)/(0.736 * (4 \pi * 0.5^(2)) * 0.8)}\n\n\nv_(t)'=10.30 \;\rm m/s$

Thus, we can conclude that the value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.

Learn more about the terminal velocity here:

brainly.com/question/2654450

Answer:

The terminal velocity at sea level is 7.99 m/s

The terminal velocity at an altitude of 5000 m is 10.298 m/s

Explanation:

mass of sphere m = 10 kg

radius of sphere r = 0.5 m

air density at sea level p = 1.22 kg/m^3

drag coefficient Cd = 0.8

terminal velocity = ?

Area of the sphere A = $4\pi r^(2)$ = 4 x 3.142 x $0.5^(2)$ = 3.142 m^2

terminal velocity is gotten from the relationship

$Vt = \sqrt{(2mg)/(pACd) }$

where g = acceleration due to gravity = 9.81 m/s^2

imputing values into the equation

$Vt = \sqrt{(2*10*9.81)/(1.22*3.142*0.8) }$ = 7.99 m/s

If at an altitude of 5000 m where air density = 0.736 kg/m^3, then we replace value of air density in the relationship as 0.736 kg/m^3

$Vt = \sqrt{(2mg)/(pACd) }$

$Vt = \sqrt{(2*10*9.81)/(0.736*3.142*0.8) }$ = 10.298 m/s

At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the tires rotating? Part (b) What is the centripetal acceleration at the edge of the tire in m/s^2?

Answers

Answer:

a) Revolutions per minute = 2.33

b) Centripetal acceleration = 11649.44 m/s²

Explanation:

a) Angular velocity is the ratio of linear velocity and radius.

Here linear velocity = 72 m/s

Radius, r = 0.89 x 0. 5 = 0.445 m

Angular velocity

$\omega =(72)/(0.445)=161.8rad/s$

Frequency

$f=(2\pi)/(\omega)=(2* \pi)/(161.8)=0.0388rev/s=2.33rev/min$

Revolutions per minute = 2.33

b) Centripetal acceleration

$a=(v^2)/(r)$

Here linear velocity = 72 m/s

Radius, r = 0.445 m

Substituting

$a=(72^2)/(0.445)=11649.44m/s^2$

Centripetal acceleration = 11649.44m/s²

How does an increase in cold working effect Modulus of Elasticity and why?

Answers

Answer:

There is a decrease in modulus of elasticity

Explanation:

Young's Modulus of elasticity also known as elastic modulus is the deformation of a body along a particular axis under the action of opposing forces along that axis. at atomic levels, it depends on bond energy or strength.

In cold working processes, plastic deformation a metal occurs below its re-crystallization temperature due to which crystal structure of metal gets distorted and as a result of dislocations fractures also occur resulting in hardening of metal but bonds at atomic levels defining elasticity are temporarily affected.

Thus an increase in cold working results in a decrease in modulus of elasticity.

A certain lightning bolt moves 40 C of charge. How many fundamental units of charge |qe| is this?

Answers

Answer to A certain lightning bolt moves 40.0 C of charge. How many fundamental units of charge | qe | is this? . ... charge, N is the total number of electron or protons that constitute total charge Q.

A 0.010 kg ball is shot from theplunger of a pinball machine.Because of a centripetal force of0.025 N, the ball follows a
circulararc whose radius is 0.29 m. What isthe speed of the
ball?

Answers

Answer:

v = 0.85 m/s

Explanation:

Given that,

Mass of the ball, m = 0.01 kg

Centripetal force on the ball, F = 0.025 N

Radius of the circular path, r = 0.29 m

Let v is the speed of the ball. The centripetal force of the ball is given by :

$F=(mv^2)/(r)$

$v=\sqrt{(Fr)/(m)}$

$v=\sqrt{(0.025* 0.29)/(0.01)}$

v = 0.85 m/s

So, the speed of the ball is 0.85 m/s. Hence, this is the required solution.

15.Restore the battery setting to 10 V. Now change the number of loops from 4 to 3. Explain what happens to the magnitude and direction of the magnetic field. Now change to 2 loops, then to 1 loop. What do you observe the relationship to be between the magnitude of the magnetic field and the number of loops for the same current

Answers

Answer:

we see it is a linear relationship.

Explanation:

The magnetic flux is u solenoid is

B = μ₀ N/L I

where N is the number of loops, L the length and I the current

By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops

B = (μ₀ I / L) N

the amount between paracentesis constant, in the case of 4 loop the field is worth

B = cte 4

N B

4 4 cte

3 3 cte

2 2 cte

1 1 cte

as we see it is a linear relationship.

In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,

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