<span>364N should be your answer.. hope this helps
</span>
Answer:
1. t = 0.0819s
2. W = 0.25N
3. n = 36
4. y(x , t)= Acos[172x + 2730t]
Explanation:
1) The given equation is

The relationship between velocity and propagation constant is

v = 15.87m/s
Time taken, 

t = 0.0819s
2)
The velocity of transverse wave is given by


mass of string is calculated thus
mg = 0.0125N

m = 0.00128kg


0.25N
3)
The propagation constant k is

hence

0.036 m
No of wavelengths, n is

n = 36
4)
The equation of wave travelling down the string is
![y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]](https://tex.z-dn.net/?f=y%28x%2C%20t%29%3DAcos%5Bkx%20-wt%5D%5C%5C%5C%5Cbecomes%5C%5C%5C%5Cy%28x%20%2C%20t%29%3D%20Acos%5B%28172%20rad.m%29x%20%2B%20%282730%20rad.s%29t%5D)
![without, unit\\\\y(x , t)= Acos[172x + 2730t]](https://tex.z-dn.net/?f=without%2C%20unit%5C%5C%5C%5Cy%28x%20%2C%20t%29%3D%20Acos%5B172x%20%2B%202730t%5D)
Answer:
d. The magnitude of the work done by the earth on the satellite is non zero
Explanation:
The work done is equal to the product of the force and the distance moved in the direction of the force, the force and the distance act perpendicular to one another, therefore no work is done in the circular motion of the movement of the earth.
Answer:
7 m .
Explanation:
For destructive interference
Path difference = odd multiple of λ /2
Wave length of sound from each of A and B.
= speed / frequency
λ = 334 / 172 = 2 m
λ/2 = 1 m
If I am 1 m away from B , the path difference will be
8 - 1 = 7 m which is odd multiple of 1 or λ /2
So path difference becomes odd multiple of λ /2.
This is the condition of destructive interference.
So one meter is the closest distance which I can remain at so that i can hear destructive interference.