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dangina [55]
3 years ago
10

A cylinder within a piston expands from a volume of 1.00 L to a volume of 2.00 L against an external pressure of 1.00 atm. How m

uch work (in Joule) was done by the expansion?
Physics
1 answer:
UkoKoshka [18]3 years ago
6 0

Answer:

1.671L

Explanation:

In the given case the pressure is constant therefore it is an isobaric process, the process in which the pressure remains constant is called as isobaric process.

Work done= external pressure× change in volume.

288= 2×( final volume-intial volume)×101.32

144÷101.32=(finalvolume-0.250)

1.421=final volume-0.250

final volume=1.671L

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When an object is placed 110 cm from a diverging thin lens, its image is found to be 55 cm from the lens. The lens is removed, a
Kazeer [188]

Explanation:

Given that,

Object distance u= -110 cm

Image distance v= 55 cm

We need to calculate the focal length for diverging lens

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{-f}=\dfrac{1}{55}-\dfrac{1}{-110}

\dfrac{1}{f}=-\dfrac{3}{110}

f=-36.6\ cm

The focal length of the diverging lens is 36.6 cm.

Now given a thin lens with same magnitude of focal length 36.6 cm is replaced.    

Here, The object distance is again the same.

We need to calculate the image distance for converging lens

Using formula of lens

\dfrac{1}{36.6}=\dfrac{1}{v}-\dfrac{1}{-110}

Here, focal length is positive for converging lens

\dfrac{1}{v}=\dfrac{1}{36.6}-\dfrac{1}{110}

\dfrac{1}{v}=\dfrac{367}{20130}

v=54.85\ cm

The distance of the image is 54.85 cm from converging lens.

Hence, This is the required solution.

5 0
3 years ago
To date, Saturn is known to have 13 rings made from which of the following?
ankoles [38]

I think the answer is Dust. Moons and stars definitely don't seem likely and dark particles, I am not even sure what those are. But I have seen rings on other planets before. Hope this helps. :)

5 0
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NISA [10]

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An object has a mass of 7 kg and is accelerating at 4 m/s2. How far would the object move if it took 168 J of work to move it?​
KIM [24]

Answer:

6m

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6 0
3 years ago
A 2.30-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. a solid sph
Marina86 [1]

Solution:


initial sphere mvr = final sphere mvr + Iω 
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m² 
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω 
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So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)² 
E = 12.64 J becomes PE = mgh, so 
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h = 0.29 m 

h = L(1 - cosΘ) → where here L is the distance to the CM 
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Θ = arccos((1-0.29)/1) = 44.77 º 

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