For a home sound system, two small speakers are located so that one is 52 cm closer to the listener than the other. for what value of the fundamental frequency of audible sound will these speakers produce constructive interference at the listener? the speed of sound is 344 m/sec. (1 point)

We want a sound wave with a wavelength of 0.52 meters or a natural fraction thereof. We'll work in MKS.

w = 0.52/n

That's length.  We have speed 344 meters/second so w corresponds to a frequency of

f = 344 / w = n (344/.52)

f = 661.5 n  Hertz

I don't really agree with how they're saying it, but all the fundamental talk is probably trying to tell us n=1,

661.5 Hertz

Any multiple of that will also produce constructive interference; we can go to about n=30 before we're out of the audio range.

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s