How much energy does the light bulb dissipate in 1 minmin when half that voltage is supplied? Express your answer in kilojoules to two significant figures.

E = 15 P₀

Explanation:

The power dissipated in a light bulb is

       P = V I

       V = I R

       P = V² / R

the power is defined by

       P = W / t

work equals energy

       P = E / t

we substitute

       V² / R = E / t

        E = V² t / R

let's reduce the time to SI units

         t = 1 min = 60 s

let's calculate the dissipated energy

In the exercise it does not indicate the nominal voltage of the bulb, but in general this voltage is V₀= 120 V

The applied voltage is half the nominal voltage

      V = V₀ / 2

      V = 120/2 = 60 V

       E = (V₀ / 2)² t / R

       E = ¼ t   V₀² / R

       E = ¼ 60 P₀

       E = 15 P₀

Many times the nominal power (P₀) is written on the box of the bulb

ι вєℓιєνє ιт ιѕ тнє fιяѕт ¢нσι¢є(α.)

нє ѕнσυℓ∂ ѕмιℓє ρσℓιтℓу αи∂ ѕρєαк тσ тнє ρєяѕσи ωнσ ιѕ ιитєяяυριиg αfтєя тнє ∂ιѕ¢υѕѕισи

нσρє ι ¢συℓ∂ нєℓρ.

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.