Astretched cable of length 2 m has a fundamental frequency of 6,000 hz. 9.1 the frequency of 1st mode is: (rad/s) a)-37699.1 b)-6000 c)-975.67 d)-0.9887 e)-9.2
the frequency of 3rd mode is: (rad/s) a)-97687.5 b)-113097.3 c)-0.8887 d)-159050.3

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

The chemical reactions lab we did is typically called a "copper plating lab,  because the acid solution cleaned the pennies and made the copper look shinier;   the acid solution cleaned the paperclips and made them look like pennies;   &   the copper from the pennies was removed, and then added to the paperclips.