17. A microphone of surface area 2.0 cm’ absorbs 1.1 mW of sound. What is the intensity of sound hitting the microphone?
A) 2.2 x 10-5 W/m2
B) 0.55 W/m²
C) 2.2 W/m2
D 2.8 W/m2
E) 5.5 W/m2

The intensity  of sound hitting the microphone is equal to: E. 5.5

Given the following data:

Surface area = 2.0

Power = 1.1 milliwatt to watt = 0.0011 watt.

Conversion

1 = 0.0001

2.0 = 0.0002

To determine the intensity  of sound hitting the microphone:

Mathematically, the intensity of a sound wave is given by the formula:

Where:

Substituting the given parameters into the formula, we have;

Intensity, I = 5.5

Read more on intensity here: link

Explanation:

Given that,

A microphone of surface area 2.0 cm’ absorbs 1.1 mW of sound.

Surfrace area, A = 2 cm = 0.02 m

Power absorbed, P = 1.1 mW = 0.0011 W

We need to find the intensity of sound hitting the microphone. Power per unit area is called the intensity of sound. It can be given by :

So, the intensity of sound is . Hence, the correct option is (b).

show the wavelengths of a wave. show the crests of a wave.

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.