During a typical workday (eight hours), the average sound intensity arriving at

Question

During a typical workday (eight hours), the average sound intensity arriving at Larry's ear is 1.8 x 10^-5 W/m^2. If the area of Larry's ear through which the sound passes is 2.1 times 10^-3 m^2, what is the total energy entering each of Larry's ears during the workday? The period of a simple pendulum in a grandfather clock on another planet is 1.66 s. What is the acceleration due to gravity on this planet? Assume that the length of the pendulum is 1.00 m. A mass m = 8.0 kg is attached to a spring and allowed to hang in the Earth's gravitational field. The spring stretches 3.6 cm from its equilibrium position. If allowed to oscillate, what would be its frequency? A performer seated on a trapeze is swinging back and forth with a period of 8.85 s. If she stands up, thus raising the center of mass of the trapeze + performer system by 35.0 cm, what will be the new period of the system? Treat trapeze performer as a simple pendulum. The sound intensity level is reported in decibels. If sound intensity level is reported in decibels. If one doubles the intensity of sound, by what factor does the perceived loudness, in decibels, change? Two boys are whispering in the library. The radiated sound power from one boy's mouth is 1.2 times 10^-9 W; and it spreads out uniformly in all directions. What is the minimum distance the boys must be away from the librarian so that she will not be able to hear them? The threshold of hearing for the librarian is 1.00 times 10^-12 W/m^2. As a train pulls out of the station going 50 m/s it blasts its horn, what is the frequency heard by the train if the passengers still at the station are hearing 384 Hz?

Explanation / Answer

here,

6)

time , t = 8 hour

t = 8 * 3600 = 28800 s

the power , P = 1.8 * 10^-5 W/m^2

the area of Larry's ear , area = 2.1 * 10^-3 m^2

the total energy entering each of Larry's ear , TE = P * area * t

TE = 2.1 * 10^-3 * 1.8 * 10^-5 * 28800 J

TE = 1.09 * 10^-3 J

the total energy entering each of Larry's ear is 1.09 * 10^-3 J

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