A concert loudspeaker suspended high off the ground emits 26.0 W of sound power. A small microphone with a 1.00 cm2 area is 53.0 m from the speaker. Part A What is the sound intensity at the position of the microphone

the sound intensity at the position of the microphone is 7.4 × 10⁻⁴ W/m²

Explanation:

Given the data in the question;

Sound power P = 26.0 W

Area of microphone A = 1.00 cm²

Radius r = 53.0 m

sound intensity at the position of the microphone = ?

Now, intensity at the position of the microphone can be determined using the following expression;

= P / 4πr²

We substitute

= 26.0 / ( 4 × π × (53.0 )² )

= 26.0 / ( 4 × π × 2809 )

= 26.0 / 35298.935

= 26.0 / ( 4 × π × (53.0 )² )

= 0.000736566

= 7.4 × 10⁻⁴ W/m²

Therefore, the sound intensity at the position of the microphone is 7.4 × 10⁻⁴ W/m²

frequency or wavelenth of em radiation in a vacuum

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form