The first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.
<h3>
Velocity of the wave</h3>
The velocity of the wave is calculated as follows;
v = √T/μ
where;
- T is tension
- μ is mass per unit length = 2 g/m = 0.002 kg/m
v = √(50/0.002)
v = 158.1 m/s
<h3>First harmonic or fundamental frequency of the wave</h3>
f₀ = v/λ
where;
f₀ = v/2L
f₀ = 158.1/(2 x 0.6)
f₀ = 131.8 Hz
<h3>Second harmonic of the wave</h3>
f₁ = 2f₀
f₁ = 2(131.8 Hz)
f₁ = 263.6 Hz
<h3>Third harmonic of the wave</h3>
f₂ = 3f₀
f₂ = 3(131.8 Hz)
f₂ = 395.4 Hz
Thus, the first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.
Learn more about harmonics here: brainly.com/question/4290297
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Answer:
Torque, 
Explanation:
It is given that,
Length of the wrench, l = 0.5 m
Force acting on the wrench, F = 80 N
The force is acting upward at an angle of 60.0° with respect to a line from the bolt through the end of the wrench. We need to find the torque is applied to the nut. We know that torque acting on an object is equal to the cross product of force and distance. It is given by :



So, the torque is applied to the nut is 34.6 N.m. Hence, this is the required solution.
Answer:
Total mass of combination = 2+3+5 = 10kg.
Acceleration produced = 2m/s^2
hence force =( total mass × acceleration)= (2×10)= 20 N.
Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N
applied force on 2 kg block = 20N
Force between 2 kg and 3 kg block = (20-4) = 16N. ans
Net force on 3 kg block = 3 × 2 =6N.
Applied force on 3 kg block due to 2 kg block = 16N.
hence, force between 3 kg and 5 kg block = (16-6) = 10N .
answers:-
(a) 20 N
(b) 16N
(c) 10 N
Answer:

Explanation:
The rotation rate of the man is:



The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:
![(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega](https://tex.z-dn.net/?f=%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%5Ccdot%20%280.16%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20%29%20%3D%20%5B%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%2B20000%5C%2Ckg%5Ccdot%20m%5E%7B2%7D%5D%5Ccdot%20%5Comega)
The final angular speed is:
