Water is boiling in a12-cm-deep pan with an outer diameter of 25 cm that is placed on top of a stove. The ambient air and the su
rrounding surfaces are at a temperature of 25°C, and the emissivity of the outer surface of the pan is 0.80. Assuming the entire pan to be at an average temperature of 98°C, determine the rate of heat loss from the cylindrical side surface of the pan to the surroundings by (a) natural convection and (b) radiation. (c) If water is boiling at a rate of 1.5 kg/h at 100°C, determine the ratio of the heat lost from the side surfaces of t
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Answer:
U = 0.413 J
Explanation:
the potential energy between two charges q1 and q2 is given by the following formula:
(1)
k: Coulomb's constant = 8.98*10^9 NM^2/C^2
q1: first charge = 4.6 μC = 4.6*10^-6 C
q2: second charge = 1.0 μC*10^-6 C
r: distance between charges = 10.0 cm = 0.10 m
You replace the values of all variables in the equation (1):
Hence, the energy between charges is 0.413 J