The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.
Mathematically, the frequency of the vibration of a string can be expressed as
Where,
L = Vibrating length string
T = Tension in the string
Linear mass density
At the same time we have the expression for the number of beats described as
Where
= First frequency
= Second frequency
From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:
If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well
Replacing 
for n and 202Hz for 
The frequency of the tightened is 205Hz
Answer:
83.3kg
Explanation:
GPE = m × g × h
GPE = mass of leopard × 10 × 36m
29988J = 360 × mass
mass = 83.3kg
Answer:
Frequency, 
Explanation:
Visible red light has a wavelength of 680 nanometers (6.8 x 10⁻⁷ m). The speed of light is 3.0 x 10 ⁸ m / s. What is the frequency of visible red light?
It is given that,
Wavelength of a visible red light is, 
Speed of light is, 
We need to find the frequency of visible red light. It can be calculated using below relation.
So, the frequency of visible red light is 
.
Answer:
H_w = 2.129 m
Explanation:
given,
Width of the weir, B = 1.2 m
Depth of the upstream weir, y = 2.5 m
Discharge, Q = 0.5 m³/s
Weir coefficient, C_w = 1.84 m
Now, calculating the water head over the weir
now, level of weir on the channel
H_w = y - H
H_w = 2.5 - 0.371
H_w = 2.129 m
Height at which weir should place is equal to 2.129 m.
