Recent content by fender5791

Thank you, got it now.

YEs, and I tried that. I did the limit as k went to infinity of 2^(k-1) / 2^(k+1) which I think is one, then the last bit is zero, so the total went to zero which doesn't help me much.

No sorry, the first one is outside, then subtract the quantity starting with 2^k-1

Okay, there's two questions, actually. First, determine if the series converges. SUM: (n-2)/(n^2-4n) (from n=5 to infinity) I used the integral test, found the integral to be 1/2 log(n^2-4n) from x=5 to x=t as t approaches infinity. That turned out to go to infinity so the series...

Got it, thanks.

So would k be (740N)/(.0005m) = 1.48x10^6 N/m?

If you stand on a bathroom scale, the spring inside the scale compresses 0.50 mm, and it tells you your weight is 740 N. Now if you jump on the scale from a height of 1.2 m, what does the scale read at its peak? I'm guessing you use 1/2kx^2 but otherwise, I'm not sure.

A .18 kg block sits against a horizontal spring. A 22 N force is applied by a hand and the spring is compressed .18 m. There is a coefficient of kinetic friction between the table and block of .3. Once the spring is compressed, the force holds the block in place without motion. What is k (spring...