Answer:
A) Although the speed is the same, the direction has changed. Therefore, the velocity has changed.
Explanation:
Answer:
Explanation: P = 300 W and t = 2 min = 120 s
Energy Q = Pt = 300 W · 120 s = 36 000 J.
Thus, plate can not produce 45 000 J heat.
Answer:
The maximum energy stored in the combination is 0.0466Joules
Explanation:
The question is incomplete. Here is the complete question.
Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Energy stored in a capacitor is expressed as E = 1/2CtV² where
Ct is the total effective capacitance
V is the supply voltage
Since the capacitors are connected in series.
1/Ct = 1/C1+1/C2+1/C3
Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF
1/Ct = 1/11.7 + 1/21.0 + 1/28.8
1/Ct = 0.0855+0.0476+0.0347
1/Ct = 0.1678
Ct = 1/0.1678
Ct = 5.96μF
Ct = 5.96×10^-6F
Since V = 125V
E = 1/2(5.96×10^-6)(125)²
E = 0.0466Joules
The force is opposite to the displacement
Explanation:
The electric field at a distance r from the charged particle is given by :
k is electrostatic constant
if r = 2 m, electric field is given by :
If r = 1 m, electric field is given by :
Dividing equation (1) and (2) we get :
So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.
