Answer:
Explanation:
Capacitor of 0.75μF, charged to 70V and connect in series with 55Ω and 140 Ω to discharge.
Energy dissipates in 55Ω resistor is given by V²/R
Since the 55ohms and 140ohms l discharge the capacitor fully, the voltage will be zero volts and this voltage will be shared by the resistor in ratio.
So for 55ohms, using voltage divider rule
V=R1/(R1+R2) ×Vt
V=55/(55+140) ×70
V=19.74Volts is across the 55ohms resistor.
Then, energy loss will be
E=V²/R
E=19.74²/55
E=7.09J
7.09J of heat is dissipated by the 55ohms resistor
Answer:
W = 1493.9 J = 1.49 KJ
Explanation:
The work done by the elevator on the object will be equal to the gain in is potential energy:
W = ΔP.E
W = mgΔh
where,
W = Work = ?
m = mass of object = 7.4 kg
g = 9.8 m/s²
Δh = gain in height = 20.6 m
Therefore,
W = (7.4 kg)(9.8 m/s²)(20.6 m)
<u>W = 1493.9 J = 1.49 KJ</u>
Answer:
30 V
Explanation:
Given that:
The uniform electric field = 50 N/C
Voltage = 80 V
distance = 1.0 m
The potential difference of the electric field = Δ V
E_d = V₁ - V₂
50 × 1 = 80V - V₂
50 - 80 V = - V₂
-30 V = - V₂
V₂ = 30 V