Answer:
0.7 mJ
Explanation:
<u>Identify the unknown: </u>
The work required to turn the dial from 180° to 0°
<u>List the Knowns: </u>
Capacitance when the dial is set at 180°: C = 350 pF = 350 x 10^-12 F Capacitance when the dial is set at 0°: C = 100 pF = 100 x 10^-12 F
Voltage of the battery: V = 130 V
<u>Set Up the Problem:</u>
<em><u>Energy stored in a capacitor: </u></em>
U_c=1/2*V^2*C
=1/2*Q^2/C
<em><u>When the dial is set at 180°:</u></em><em> </em>
U_c=1/2*(130)^2*350*10^-12=10^-4
Q=√2*U_c*C=4*10^-7
<u><em>When the dial is set at 0°:</em></u>
U_c=1/2*(4*10^-7)^2/100*10^-12
=8*10^-4 J
<u><em>Solve the Problem: </em></u>
ΔU_c=7*10^-4 J
=0.7 mJ
note:
there maybe error in calculation but method is correct
Answer:
the hot bulb will have high resistance to the flow of current. While the cold bulb will have a low resistance to the flow of current.
Explanation:
A conductor that does not obey Ohm's law is described as non - ohmic. An example is a filament lamp. It glows as the current passes through it.
How does the resistance of the light bulbs differ when the bulbs are cold and when the bulbs are hot ?
The resistance of the light bulbs increase gradually as its temperature is increased.
So, the hot bulb will have high resistance to the flow of current. While the cold bulb will have a low resistance to the flow of current.
Because the resistance of an impure metal wire is greater than the resistance of a pure metal wire of the same dimension.
If you, for example, poured it onto a wide cup with a volume equal to the total volume of the sand particles, the sand would not spread out to fill the container but would bunch up together in the middle.
Answer:
The correct answer would be A
Answer:
Yes, both of them are correct.
