A typical incandescent light bulb consumes 75 W of power and has a mass of 30 g. You want to save electrical energy by dropping the bulb from a height great enough so that the kinetic energy of the bulb when it reaches the floor will be the same as the energy it took to keep the bulb on for 1.0 hour. From what height should you drop the bulb, assuming no air resistance and constant g?

The height will be 917431.2 m.

Explanation:

Power of bulb = 75 W

Time kept on 1 hr = 60 x 60 = 3600 sec

Energy of bulb = power x time

E = 75 x 3600 = 270000 J

From conservation of energy, kinetic energy of the bulb is equal to the potential energy of the bulb due to its height of fall.

Potential energy = m x g x h

Where g = acceleration due to gravity 9.81 m/s2

m = mass = 30 g = 0.03 kg

PE = 0.03 x 9.81 x h = 0.2943h

Equating withe the energy of bulb (still obeying energy conservation)

270000 = 0.2943h

h = 270000/0.2943 = 917431.2 m

0.11 m/s and 0.28 m/s should be correct

when a device is used in a circuit in which the voltage is 81 v, the current flowing through the device is 3 a. what is the resistance of the device?

a. 27 ω

b. 40.5 a

c. 243 ω

d. 243 a

the correct answer is a. 27