Apipe of length 6.8 m is closed at one end and sustains a standing wave at its second overtone. determine the distance between a node and an adjacent antinode.

For a pipe closed at one end, the second overtone is the 3rd natural frequency as shown in the diagram.

For the fundamental frequency,
λ = 4L
where
λ = wavelength
L = the length of the tube

For the second overtone (3rd frequency)
λ = (4/3)L

Because L = 6.8 m, therefore
λ = (4/3)*(6.8 m) = 9.0667 m

The distance between a node and an adjacent antinode is λ/2, which is
9.0667/2 = 4.533 m

4.5 m

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j