Explanation:
The following measures can be taken to prevent or control high temperature caused by thermal pollution:
1. Heated water from the industries can treated before discharging directly to the water bodies.
2. Heated water from the industries can be treated by the installation of cooling ponds and cooling towers.
3. Industrial treated water can be recycled for domestic use or industrial heating.
4. Through artificial lakes: In this lake Industries can discharge their used or heated water at one end and water for cooling purposes may be withdrawn from the other end. The heat is eventually dissipated through evaporation.
(if it's frictionless the length doesn't even matter :) )
It would have the same kinetic energy down as the potential energy up. That is,
or
(the mass doesn't even matter). The result is
, so only the height matters really. It is almost 9 (it is
).
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
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