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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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7. A scientist studying a squid observes that the squid at rest
Rzqust [24]

The average force on the squid during the ejection of 0.60 kg of water at a velocity of 15.0 m/s in 0.15 seconds is 60 N.              

We can calculate the average force with the average acceleration as follows:

F = m\overline{a}   (1)

Where:

  • m: is the mass of water = 0.60 kg
  • \overline{a}: is the average acceleration

The <em>average acceleration</em> is given by the change of velocity in an interval of time

\overline{a} = \frac{v_{f} - v_{i}}{t_{f} - t_{i}}   (2)

Where:

  • v_{i}: is the initial velocity = 0 (the squid is at rest)
  • v_{f}: is the final velocity = 15.0 m/s
  • t_{i}: is the initial time = 0  
  • t_{f}: is the final time = 0.15 s

Now we can find the <em>average force</em> after entering equation (2) into (1)

F = m(\frac{v_{f} - v_{i}}{t_{f} - t_{i}}) = 0.60 kg(\frac{15.0 m/s - 0}{0.15 s}) = 60 N  

Therefore, the average force on the squid during the propulsion is 60 N.

Find more about average force here:

  • brainly.com/question/20902034
  • brainly.com/question/12916904

I hope it helps you!

3 0
3 years ago
Determine the angular velocity ω of the telescope as it orbits around the Sun.
lara31 [8.8K]
The JWST is postioned about 1.5 million kilometers from the earth on the side facing away from the sun
5 0
3 years ago
Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?
Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0
3 years ago
To ancient peoples, why were planets special?
coldgirl [10]

Answer:

Planets were like gods.

Explanation:

To the people of many ancient civilizations, the planets were thought to be deities. Our names for the planets are the Roman names for these deities. For example, Mars was the god of war and Venus the goddess of love.

7 0
3 years ago
A student (m = 68 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a
Gnesinka [82]

Answer:

5.7141 m

Explanation:

Here the potential and kinetic energy will balance each other

mgh=\frac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}

This is the initial velocity of the system and the final velocity is 0

t = Time taken = 0.04 seconds

F = Force = 18000 N

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

v=u+at\\\Rightarrow a=\frac{v-u}{t}

From Newton's second law

F=ma\\\Rightarrow F=m\frac{v-u}{t}\\\Rightarrow 18000=68\frac{0-\sqrt{2gh}}{0.04}\\\Rightarrow \frac{18000}{68}\times 0.04=-\sqrt{2\times 9.81\times h}\\\Rightarrow 10.58823=-\sqrt{2\times 9.81\times h}

Squarring both sides

112.11061=2\times 9.81\times h\\\Rightarrow h=\frac{112.11061}{2\times 9.81}\\\Rightarrow h=5.7141\ m

The height from which the student fell is 5.7141 m

5 0
3 years ago
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