Answer:
Answer:
Speed of the wave in the string will be 3.2 m/sec
Explanation:
We have given frequency in the string fixed at both ends is 80 Hz
Distance between adjacent antipodes is 20 cm
We know that distance between two adjacent anti nodes is equal to half of the wavelength
So \frac{\lambda }{2}=20cm
2
λ
=20cm
\lambda =40cmλ=40cm
We have to find the speed of the wave in the string
Speed is equal to v=\lambda f=0.04\times 80=3.2m/secv=λf=0.04×80=3.2m/sec
So speed of the wave in the string will be 3.2 m/sec
Your answer should be 16.3 m 79.4º east of north
It is called condensation. Hope this helped!
We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.
1) Equilibrium of forces:

where

is the weight of the person

is the weight of the scaffold
Re-arranging, we can write the equation as

(1)
2) Equilibrium of torques:

where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using

and replacing T1 with (1), we find

from which we find

And then, substituting T2 into (1), we find
<span>Work is required to pull a nucleon out of an atomic nucleus. It has more mass outside the nucleus.</span>