Parallel-plate capacitor has there fore formula is
<span>C=(<span>ϵ0</span>A)/d
putting values</span>C=(8.85*10^-12*pi*.05^2)/.00063
=1.1*10^-10F
then Q=CV=1.1*10^-10*1000=1.1*10^-7C
as
<span>η=Q/A</span><span>therefore
(1.1*10^-7)/(pi*.05^2)
=1.4*10^-5C/m^ our answer
hope this helps</span>
Answer:
7.72 Liters
Explanation:
normal body temperature = T_body =37° C
temperature of ice water = T_ice =0°c
specfic heat of water = c_{water} =4186J/kg.°C
if the person drink 1 liter of cold water mass of water is = m = 1.0kg
heat lost by body is Qwater =mc_{water} ΔT
= mc{water} ( T_ice - T_body)
= 1.0×4186× (0 -37)
= -154.882 ×10^3 J
here negative sign indicates the energy lost by body in metabolic process energy expended due to brisk - hour long walk is Q_{walk} = 286 kilocalories
= 286×4186J
so number of liters of ice water have to drink is
n×Q_{water} =Q_{walk} n= Q_{walk}/ Q_{water}
= 286×4186J/154.882×10^3 J
= 7.72 Liters
