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)
Lysosomes are used by the cell to digest or breakdown multifaceted organic molecules
Answer:
0.15 s
Explanation:
From the question given above, the following data were obtained:
Speed of sound (v) = 330 m/s
Distance (x) = 25 m
Time (t) =?
The time taken for the echo of the sound to the bat can be obtained as follow:
v = 2x / t
330 = 2 × 25 / t
330 = 50 / t
Cross multiply
330 × t = 50
Divide both side by 330
t = 50 / 330
t = 0.15 s
Thus, it will take 0.15 s for the echo of the sound to the bat
Answer:
14. scolded
15. supporter
16. i am not sure but I think it is removes
17. comfortable
18. outdated
19. drink
20. it can be fantastic or wealthy
