Answer:
Explanation:
Initial separation of plate = d
final separation = 2d
The capacitance of the capacitor will reduce from C to C/2 because
capacitance = ε A / d
d is distance between plates.
As the batteries are disconnected , charge on the capacitor becomes fixed .
Initial charge on the capacitor
= Capacitance x potential difference
Q = C ΔV
Final charge will remain unchanged
Final charge = C ΔV
Final capacitance = C/2
Final potential difference = charge / capacitance
= C ΔV / C/2
= 2 ΔV
Potential difference is doubled after the pates are further separated.
Answer:
Vmax=11.53 m/s
Explanation:
from conservation of energy

Spring potential energy =potential energy due to elevation
0.5*k*x²= mg
=mgh
0.5*k*2.3²= 430*9.81*6
k=9568.92 N/m
For safety reason
k"=1.13 *k= 1.13*9568.92
k"=10812.88 N/m
agsin from conservation of energy

spring potential energy=change in kinetic energy
0.5*k"*x²=0.5*m*
10812.88 *2.3²=430*
=11.53 m/s
One form of Ohm's Law says . . . . . Resistance = Voltage / Current .
R = V / I
R = (12 v) / (0.025 A)
R = (12 / 0.025) (V/I)
<em>R = 480 Ohms</em>
I don't know if the current in the bulb is steady, because I don't know what a car's "accumulator" is. (Floogle isn't sure either.)
If you're referring to the car's battery, then the current is quite steady, because the battery is a purely DC storage container.
If you're referring to the car's "alternator" ... the thing that generates electrical energy in a car to keep the battery charged ... then the current is pulsating DC, because that's the form of the alternator's output.