Answer:
The work and heat transfer for this process is = 270.588 kJ
Explanation:
Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k
The Pressure-Volume relation is <em>PV</em> = <em>C</em>
<em>T = C </em> for isothermal process
Calculating for the work done in isothermal process
<em>W</em> = <em>P</em>₁<em>V</em>₁ ![ln[\frac{P_{1} }{P_{2} }]](https://tex.z-dn.net/?f=ln%5B%5Cfrac%7BP_%7B1%7D%20%7D%7BP_%7B2%7D%20%7D%5D)
= <em>mRT</em>₁ [∵<em>pV</em> = <em>mRT</em>]
= (5) (0.287) (272.039) ![ln[\frac{2.0}{1.0}]](https://tex.z-dn.net/?f=ln%5B%5Cfrac%7B2.0%7D%7B1.0%7D%5D)
= 270.588 kJ
Since the process is isothermal, Internal energy change is zero
Δ<em>U</em> = 
From 1st law of thermodynamics
Q = Δ<em>U </em>+ <em>W</em>
= 0 + 270.588
= 270.588 kJ
Answer:
charge on each
Q1 = 2.06 ×
C
Q2 = 7.23 × C
when force were attractive
Q1 = 1.07 × 
C
Q2 = -1.39 × C
Explanation:
given data
total charge = 93.0 μC
apart distance r = 1.14 m
force exerted F = 10.3 N
to find out
What is the charge on each and What if the force were attractive
solution
we know that force is repulsive mean both sphere have same charge
so total charge on two non conducting sphere is
Q1 + Q2 = 93.0 μC = 93 ×
C
and
According to Coulomb's law force between two sphere is
Force F = 
.........1
Q1Q2 = 
here F is force and r is apart distance and k is 9 × 
N-m²/C² put all value we get
Q1Q2 = 
Q1Q2 = 1.49 × 
C²
and
we have Q2 = 93 × C - Q1
put here value
Q1² - 93 × Q1 + 1.49 × = 0
solve we get
Q1 = 2.06 × C
and
Q1Q2 = 1.49 ×
2.06 × Q2 = 1.49 ×
Q2 = 7.23 × C
and
if force is attractive we get here
Q1Q2 = - 1.49 × C²
then
Q1² - 93 × Q1 - 1.49 × = 0
we get here
Q1 = 1.07 × C
and
Q1Q2 = - 1.49 ×
2.06 × Q2 = - 1.49 ×
Q2 = -1.39 × C
Answer:
The tube should be held vertically, perpendicular to the ground.
Explanation:
As the power lines of ground are equal, so its electrical field is perpendicular to the ground and the equipotential surface is cylindrical. Therefore, if we put the position fluorescent tube parallel to the ground so the both ends of the tube lie on the same equipotential surface and the difference is zero when its potential.
And the ends of the tube must be on separate equipotential surfaces to optimize potential. The surface near the power line has a greater potential value and the surface farther from the line has a lower potential value, so the tube must be placed perpendicular to the floor to maximize the potential difference.
If a coastline has a very unusual shape it's normally caused by either a flood or dam.
hope this helps!
