Answer:
(a) Wavelength is 0.436 m
(b) Length is 0.872 m
(c) 11.518 m/s
Solution:
As per the question:
The eqn of the displacement is given by:
(1)
n = 4
Now,
We know the standard eqn is given by:
(2)
Now, on comparing eqn (1) and (2):
A = 1.22 cm
K = 

where
A = Amplitude
K = Propagation constant
= angular velocity
Now, to calculate the string's wavelength,
(a) 
where
K = propagation vector


(b) The length of the string is given by:


(c) Now, we first find the frequency of the wave:



Now,
Speed of the wave is given by:


Answer:
Acceleration=24.9ft^2/s^2
Angular acceleration=1.47rads/s
Explanation:
Note before the ladder is inclined at 30° to the horizontal with a length of 16ft
Hence angular velocity = 6/8=0.75rad/s
acceleration Ab=Aa +(Ab/a)+(Ab/a)t
4+0.75^2*16+a*16
0=0.75^2*16cos30°-a*16sin30°---1
Ab=0+0.75^2sin30°+a*16cos30°----2
Solving equation 1
(0.75^2*16cos30/16sin30)=angular acceleration=a=1.47rad/s
Also from equation 2
Ab=0.75^2*16sin30+1.47*16cos30=24.9ft^2/s^2
Answer:
C because it is in earth's mantle
Answer:
The wavelength is 3500 nm.
Explanation:
d= 
n= 1
θ= 30°
λ= unknown
Solution:
d sinθ = nλ
λ = 
λ = 3500 nm