When the heat of the sun shines on the water<span> in oceans, lakes, rivers and streams, the </span>water<span> evaporates, rising up into the air as </span>water<span> vapor. As it moves higher into the sky, it cools.
</span>
The process you're fishing for is "polarization", but that's a
misleading description.
Polarization doesn't do anything to change the light waves.
It simply filters out (absorbs, as with a polarizing filter) the
light waves that aren't vibrating in the desired plane, and
allows only those that are to pass.
The intensity of a light beam is always reduced after
polarizing it, because much (most) of the original light
has been removed.
A laser light source may be thought of as an exception,
since everything coming out of the laser is polarized.
Explanation:
<em>Here </em><em>it </em><em>is </em><em>given </em>
<em>Work </em><em>(</em><em>W) </em><em> </em><em>=</em><em> </em><em>3</em><em>5</em><em>6</em><em>0</em><em> </em><em>J</em>
<em>Time </em><em>(</em><em>t) </em><em> </em><em>=</em><em> </em><em>5</em><em>5</em><em> </em><em>sec</em>
<em>power </em><em>(</em><em>P) </em><em> </em><em>=</em><em> </em><em>?</em>
<em>We </em><em>know </em><em>we </em><em>have </em><em>the </em><em>formula </em>
<em>
</em>
<em>P </em><em>=</em><em> </em><em>3</em><em>5</em><em>6</em><em>0</em><em>/</em><em>5</em><em>5</em>
<em>P </em><em>=</em><em> </em><em>6</em><em>4</em><em>.</em><em>7</em><em>3</em><em> </em><em>watt</em>
Answer:
(a) Since net charge remains same,after immersion Q is same
(b) I. 14.56pF ii. 3.05V
(c) ΔU = 5.204nJ
Explanation:
a)
C = kεA/d
k=1 for air
ε is 8.85x10-12F/m
A = .0025m2
d = .125m
C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF
Q = CV = .177pF * 244V = 43.188pC
Since net charge remains same,after immersion Q is same
b)
C = kεA/d, for distilled water k is approx. 80
Cwater = Cair x k
=0.177pF x 80 = 14.16pF
Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V
c) Change in energy: ΔU = Uwater - Uair
Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ
Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ
ΔU = 5.204nJ
