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rosijanka [135]
3 years ago
9

How much work is required to compress a gas from a volume of 5.0 l to a volume of 2.5 l by exerting a constant external pressure

of 1.5 atm? calculate the answer in l-atm and joules?
Physics
1 answer:
ch4aika [34]3 years ago
4 0
A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs. 
Read more on Brainly.com - brainly.com/question/1581851#readmore
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A stone is dropped into a well. The sound of the splash is heard 3.5 seconds later. What is the depth of the well? Take the spee
Naddika [18.5K]

Answer:

The depth of the well, s = 54.66 m

Given:

time, t = 3.5 s

speed of sound in air, v = 343 m/s

Solution:

By using second equation of motion for the distance traveled by the stone when dropped into a well:

s = ut +\frac{1}{2}at^{2}

Since, the stone is dropped, its initial velocity, u = 0 m/s

and acceleration is due to gravity only, the above eqn can be written as:

s = \frac{1}{2}gt'^{2}

s = \frac{1}{2}9.8t^{2} = 4.9t'^{2}                     (1)

Now, when the sound inside the well travels back, the distance covered,s is given by:

s = v\times t''

s = 343\times t''                                              (2)

Now, total time taken by the sound to travel:

t = t' + t''

t'' = 3.5 - t'                                                                        (3)

Using eqn (2) and (3):

s = 343(3.5 - t')                                                                 (4)

from eqn (1) and (4):

4.9t'^{2} = 343(3.5 - t')

4.9t'^{2} + 343t' - 1200.5 = 0

Solving the above quadratic eqn:

t' = 3.34 s

Now, substituting t' = 3.34 s in eqn (2)

s = 54.66 m

3 0
3 years ago
Why can you see your own reflected image in a mirror but not on a dry, painted wall?
kaheart [24]
A wall uses diffuse reflection while a mirror uses specular reflection. For example, when parallel light rays enter a mirror, they remain parallel when exiting the mirror, allowing you to see a reflection of the light rays. On the contrary, when incident light rays enter a wall which is painted, the rays scatter, not allowing you to see anything but a painted wall. 
7 0
3 years ago
Gravitational Potential Energy = mass x gravity x height
Blizzard [7]
M= gpe / gh
G= gpe / mh
H=gpe / mg

7 0
3 years ago
Read 2 more answers
Which group of measurements is the most precise? A) 2 g, 3 g, 4 g B) 2 g, 2.5 g, 3 g C) 2.0 g, 3.0 g, 4.0 g, 5.0 g D) 2.0 g, 3.0
siniylev [52]
The answer to your question is A
3 0
3 years ago
A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di
siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

  • The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

                                  E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\

part a)

Electric Field strength at point P: a = 0.75 m

E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2  

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

 

6 0
3 years ago
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