An unknown resistor is connected between the terminals of a 3.00 v battery. energy is dissipated in the resistor at the rate of 0.560 w. the same resistor is then connected between the terminals of a 1.50 v battery. at what rate is energy now dissipated?

The energy dissipate at a rate of 11.34 W when the resistor is connected between the terminals of a 1.50 V battery.

Explanation:

the rate at which energy  dissipeted is given by power and we know that when given resistance, R and voltage, V then power is given by:

P = V^2/R

R = V^2/P

  = (3.00)^2/(0.560)

  = 5.04 ohms

then when the resistor is now connected between the terminals of a 1.50 V battery, then the power is given by:

P = V^2/R

  = (1.50)^2/(5.40)

  = 11.34 W.

Therefore, the energy dissipate at a rate of 11.34 W when the resistor is connected between the terminals of a 1.50 V battery.

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5