Final answer:
The noise from a rock concert rated at 113 decibels is approximately 7943 times more intense than the noise from a hot water pump rated at 74 decibels.
Step-by-step explanation:
The question asks to determine how many times more intense the noise from a rock concert is compared to that of a hot water pump, given their noise ratings in decibels. We can use the formula d = 10 log(P / P0) to solve this problem, where d is the decibel level, P is the power or intensity of the sound, and P0 is the reference intensity or the weakest sound the human ear can hear.
First, we calculate the difference in decibels between the rock concert and the hot water pump.
113 dB (rock concert) - 74 dB (hot water pump) = 39 dB
Each factor of 10 in intensity corresponds to 10 dB. This means that a difference of 39 dB corresponds to 10^(39/10) times the intensity. Therefore, we can calculate this as:
Intensity factor = 10^(39/10) = 10^3.9
This calculation yields an intensity factor of approximately 7943.
Thus, the rock concert's noise is 7943 times as intense as the hot water pump.