Alightbulb is connected to a 60 hz power
source having a maximum voltage of 156 v.,
what is the resistance of the light bulb that
uses an average power of 38.8 w?

The resistance of the light bulb is, R = 4.02 Ω                

Explanation:

Given data,

The frequency of the current, f = 60 Hz

The voltage of the source, V =  156 V

The power usage of the bulb, P = 38.8 watts

The resistance of the light bulb is given by the formula,

                                            R = V² / P

Substituting the values,

                                             R = 156 / 38.8

                                                 = 4.02 Ω      

Hence, the resistance of the light bulb is, R = 4.02 Ω                

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6

terd muffin is the correct answer