Asked by Angel
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.
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.
Answers
Answered by
bobpursley
period=2PIsqrt(length/g)
square both sides...
T^2=(2PI)^2 l/g
g= (2PI)^2 Length/period^2
square both sides...
T^2=(2PI)^2 l/g
g= (2PI)^2 Length/period^2
Answered by
Angel
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..
The equation I have is:
T=2(Pie)*Sqt(L/G)
I need to re-arrange that..
Answered by
bobpursley
I would square it first, as I did, then rearrange.
Answered by
Angel
By following your equation I got:
5.797880106
If I rearrange it would be:
T^2 = 2(Pie)L/G
then what?
5.797880106
If I rearrange it would be:
T^2 = 2(Pie)L/G
then what?
Answered by
bobpursley
You did not square 2PI.
T^2=(2pi)^2 l/g
g= (2PI)^2 l/t^2
T^2=(2pi)^2 l/g
g= (2PI)^2 l/t^2
Answered by
Angel
Last question.. sorry
PL = Pie?
PL = Pie?
Answered by
Angel
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)
I put
(2<Pie>.594)^2.594/(1.55^2)
Answered by
bobpursley
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.
(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.
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