The period of a simple pendulum is given by:

where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:

(1)
We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.
We know it makes N=441 oscillations in t=1090 s, therefore its frequency is

And its period is the reciprocal of its frequency:

So now we can use eq.(1) to find the gravitational acceleration of the planet:
We assign the variables: T as tension and x the angle of the string
The <span>centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = </span><span>T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx.
</span>
<span>(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. </span>
<span>(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. </span>
<span>(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. </span>
<span>(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.
</span><span>This minimum v is obtained when T=0 </span><span>and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.</span>
If you mean the SI Unit of GPE, the answer is J for Joules.
if that's not the question being asked, i would need a little more elaboration please :)
Answer:
Number of neautrons
Explanation:
Atoms from two different elements may have the same number of neutrons, but never the same number of protons. The number of protons is unique to the element and represents the atomic number. (sorcastic.org)
Answer:

Explanation:
Given data
Mass m=67.0 kg
Final Speed vf=8.00 m/s
Initial Speed vi=2.00 m/s
Distance d=25.0 m
Force F=30.0 N
From work-energy theorem we know that the work done equals the change in kinetic energy
W=ΔK=Kf-Ki=1/2mvf²-1/2mvi²
And

So

and we know that the force the sprinter exerted Fsprinter the force of the headwind Fwind=30.0N
So