What is the ratio of the displacement amplitudes of two sound waves given that they are both5.0 kHz but have a 3.0 dB intensity level difference?

The ratio of the displacement amplitudes of two sound waves is 1.16.

Explanation:

Given that,

Frequency = 5.0 kHz

Intensity level difference = 3.0 dB

We know that,

The sound intensity is inversely proportional to the square of distance.

The sound intensity for first wave is

...(I)

The sound intensity for second wave is

...(II)

We need to calculate the ratio of intensity

From equation (I) and (II)

Put the value into the formula

We need to calculate the ratio of the displacement

Using formula of displacement

Put the value into the formula

Hence, The ratio of the displacement amplitudes of two sound waves is 1.16.

i think the answer would be b

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.