Length of the pipe = 0.39 m
Number of harmonics = 3
Now there are 3 loops so here we can say


now here at the center of the pipe it will form Node
we need to find the distance of nearest antinode
So distance between node and its nearest antinode will be


So the distance will be 6.5 cm
Answer:
x = 5.29 m
Explanation:
given,
weight of stretch = 37 N
left-hand spring constant (k₁)= 2.7 N/cm
right hand spring constant(k₂)= 4.3 N/ cm
spring are connected in parallel
F = F₁ + F₂
F = k₁x + k₂x
F = (k₁+ k₂)x
37= (4.3+ 2.7)x
7 x = 37
x = 5.29 m
Answer:

Explanation:
Given that there are two force of 1 pound each at right angles to each other.
The from the vector law of addition:

where:
resultant force
be the two of the forces to be added.


Answer:
a) Diffusion coefficient, D = 1.5 in/hr
b) Mean jump frequency, f = 0.0833 Hz
Explanation:
a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:
..........(1)
Where <r> = mean displacement
D = Diffusion coefficient
t = time = 12 hrs
sum of the squares of the distance divided by 100 is 36 in2.
<r>²= 36 in²
Substituting these values into equation (1) above

b) Mean jumping distance, <r> = 0.1 inches
Applying equation (1) again
Where D = 1.5 in/hr


The mean jump frequency, f = 1/t
f = 1/12
f = 0.0833 Hz