Answer:
gravity
Explanation:
We use gravity to do different force.
Answer:
No temperature change occurs from heat transfer if ice melts and becomes liquid water (i.e., during a ... to change 1 kg of liquid water at the normal boiling point (100ºC at atmospheric pressure) to steam (water vapor).
The melting point of lead is 327.3o C. Assume the final temperature of the system is T. Then the amount of energy released by the lead as it solidifies is. ΔQ = mleadLlead = 0.09 kg*(2.45*104 J/kg) = 2205 J
C. fast-moving particles colliding with slow-moving particles.
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
Answer:
4/3 pi R^3 = pi r^2 L equating volume of sphere and wire
r = (4 R^3 / 3 * L)^1/2 solving for radius of wire
r = (4 * 6^3 / 3 * 32)^1/2
r = 9^1/2 cm = 3 cm = .03 meters