Two small identical speakers are connected (in phase) to the same source. The speakers are 3 m apart and at ear level. An observer stands at X, 4 m in front of one speaker. If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is: a. 1 m
b. 2 m
c. 3 m
d. 4 m
e. 5 m

b. 2 m

Explanation:

Given that:

the identical speakers are connected in phases ;

Let assume ; we have speaker A and speaker B which are = 3 meter apart

An observer stands at X = 4m in front of one speaker.

If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is:                  

From above;  the distance between speaker  A and speaker B can be expressed as:

The path length difference  will now be:

= 5 m - 4 m

= 1 m

Since , we are to determine the least intense sound; the destructive interference for that path length  will be half the wavelength; which is

=

= 2 m

a.   mass number from atomic number

so easy. just google the electromagnetic waves and it’ll pop up.