Answer:
P(bat) = V²r/(R+r)²
Explanation:
Let the resistance of the coil be R
Internal resistance of the battery be r
Emf of the battery = V
Power dissipated in the internal resistance of the battery is normally given as P = I²r
where I is the current flowing in the circuit.
From Ohm's law,
V = I R(eq)
R(eq) = (R + r)
I = V/(R+r)
P = I²r
P = [V/(R+r)]²r
P = V²r/(R+r)²
Hope this Helps!!!
Only within the same technology. / / /
If both of the bulbs you're comparing are incandescent, or both fluorescent, or both CFL, or both LED, then the one that uses more power is brighter. But a CFL with the same brightness as an incandescent bulb uses less power, and an LED bulb with the same brightness as both of those uses less power than either of them.
W=20 e(-kt)
A. Rearranging gives k= -(ln(w/20)/t
Substituting w= 10 and solving gives k=0.014
B. Using W=20e(-kt). After 0 hours, W=20. After 24 hours, W=14.29g. After 1 week (24x7=168h) W=1.9g
C. Rearranging gives t=-(ln(10/20)/k. Substituting w=1 and solving gives t=214 hours.
D. Differentiating gives dW/ dt = -20ke(-kt). Solving for t=100 gives dW/dt = 0.07g/h. Solving for t=1000 gives 0.0000002g/h
E. dW/dt = -20ke(-kt). But W=20e(-kt) so dW/dt = -kW
Answer:
value of heat is 18 J
2. step by step
formular w=p(volume1-volume2)
w= 1.0×10^5(0.0006-0.0004)
w= 40 J
It is the most massive planet in the solar system.
