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ruslelena [56]
3 years ago
15

During a breath, the average human lung expands by about 0.50 l. part a if this expansion occurs against an external pressure of

1.0 atm, how much work (in j) is done during the expansion?
Physics
1 answer:
saveliy_v [14]3 years ago
8 0
Work, in thermodynamics, is the amount of energy that is transferred from one system to another system without transfer of entropy. It is equal to the external pressure multiplied by the change in volume of the system. It is expressed as follows:<span>

W = PdV

Integrating and assuming that P is not affected by changes in V or it is constant, then we will have:

W = P (V2 - V1)

Substituting the given values:
P = 1.0 atm = 101325 Pa
(V2 - V1) = 0.50 L = 
W = 101325 N/m^3 ( 0.50) (1/1000) m^3
W = 50.66 N-m or 50.66 J
<span>
So, in the expansion process about 50.66 J of work is being done.
</span></span><span><span>
</span></span>
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According to our theory of solar system formation, what are asteroids and comets?
bija089 [108]
Rocks leftover from planet or moon formation.
6 0
3 years ago
Lila swam 50 meters north in 10 seconds. Find Lila's velocity
In-s [12.5K]

Explanation:

50 meters north in 10 seconds.

speed = distance ÷ time = 50 meters ÷ 10 seconds = 5m/s

8 0
3 years ago
A disc on a frictionless axle, starting from rest (0 rpm) can spin up to a rotation rate of 3820 rpm in a period of 2 seconds. (
Eddi Din [679]

Answer:

1000 Nm

2000 Nm

1.00007 seconds

Explanation:

I = Moment of inertia = 5 kgm²

\alpha = Angular acceleration

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

t = Time taken

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=5\times 200\\\Rightarrow \tau=1000\ Nm

The torque of the disc would be 1000 Nm

If \alpha=400\ rad/s^2

\tau=I\alpha\\\Rightarrow \tau=5\times 400\\\Rightarrow \tau=2000\ Nm

The torque of the disc would be 2000 Nm

From equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{3820\times \frac{2\pi}{60}-0}{400}\\\Rightarrow t=1.00007\ s

It would take 1.00007 seconds to reach 3820 rpm

6 0
4 years ago
In which list below do the experimental steps appear in the order in which they are done?
fgiga [73]
The answer is B . You begin with a purpose for the lab, then hypothesize on what you believe will happen. Next, follow the procedures. This is always the last step, Anatoly did. Reflect upon you hypothesis, did the lab support or disprove your hypothesis. Include observations you have made. Identify errors.
5 0
3 years ago
A 1.6-lb collar is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an
romanna [79]

Answer:

k = 652 lb/ft

Explanation:

Given :

Weight of the collar = 1.6 lb

The upstretched length of the spring = 6 in

Speed  = 16 ft/s

PA = 8 + 10

     = 18 inch

Let the initial elongation be $\Delta x_i$

∴ $\Delta x_i$ = 18 - 6

         = 12 inch = 1 foot

$PB = \sqrt{13^2+5^2}$

      = 13.925 inch

Final elongation in the spring

$\Delta x_B = 7.928 $ inch = 0.66 feet

Applying the conservation of the mechanical energy between A and B is

$K.E_A+P.E_{g,A}+P.E_{sp,A}= K.E_B+P.E_{g,B}+P.E_{sp,B} $

$0+mg_r+\frac{1}{2}k(\Delta x_i)^2=\frac{1}{2}mv_B^2+0+\frac{1}{2}k(\Delta x_B)^2$

$\frac{1}{2}k[(1)^2-(0.66)^2]=\frac{1.6}{2}\times (16)^2-1.6 \times 32 \times \frac{5}{12}$

0.281 \ k =204.8-21.33

k = 652 lb/ft

5 0
3 years ago
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