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

the temperature and the pressure affect solubility. temperature affects by changing the properties of the solvent. the increase in pressure increases solubility, the decrease in pressure decreases solubility.

im sorry i dont know