The best conclusion that can be drawn is that D) A current does not flow in the wire
Answer:
I think its C sorry if it's wrong
<span>if we assume the origin is at the dropping point and the object is merely dropped and not thrown up or down then y0 = 0 and v0 = 0. The equation reduces to </span>
<span>y = 0 + 0t + ½gt² </span>
<span>y = ½gt² </span>
<span>t = √(2y/g) </span>
<span>in the ft - lb - s system </span>
<span>y = -100 ft </span>
<span>g = -32.2 ft / s² </span>
<span>t = √(2y/g) </span>
<span>t = √(2(-100) / (-32.2)) </span>
<span>t = 2.5 s</span>
Answer:
The workdone is 
Explanation:
From the question we are told that
The potential difference is 
Generally the charge on 
is 
Generally the workdone is mathematically represented as
=> 
=>
