Question

What is the free-fall acceleration in a location
where the period of a 0.594 m long pendulum
is 1.55 s s?
Answer in units of m/s^2

Show equation too please.

Answer 1

period=2PIsqrt(length/g)

square both sides...

T^2=(2PI)^2 l/g

g= (2PI)^2 Length/period^2

Answer 2

I kinda have no clue what you responded.

The equation I have is:
T=2(Pie)*Sqt(L/G)

I need to re-arrange that..

Answer 3

I would square it first, as I did, then rearrange.

Answer 4

By following your equation I got:
5.797880106

If I rearrange it would be:
T^2 = 2(Pie)L/G

then what?

Answer 5

You did not square 2PI.

T^2=(2pi)^2 l/g

g= (2PI)^2 l/t^2

Answer 6

Last question.. sorry

PL = Pie?

Answer 7

I keep getting the question wrong. I think I put it wrong in my caculator.

I put

(2<Pie>.594)^2.594/(1.55^2)

Answer 8

2PI= 6.28
(2PI)^2=about 39, check it
.594=l
1.55^2= about 2.4 check it.

g=39*.594/2.4 in my head, about 9.7 check it.