Answer:
The solution and the explanation are in the Explanation section.
Explanation:
According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:
TA = W * (a * cosθ)
The torque due to effort at C point is:
TC = E * (b * cosθ)
The net torque is equal to 0, we have:
Tnet = 0
W * (a * cosθ) - E * (b * cosθ) = 0
![E=W\frac{a}{b}](https://tex.z-dn.net/?f=E%3DW%5Cfrac%7Ba%7D%7Bb%7D)
From the figure, you can observe that a/b < 1, thus E < W
( 1.05 x 10¹⁵ km ) x ( 1 LY / 9.5 x 10¹² km ) x ( 1 psc / 3.262 LY ) =
(1.05) / (9.5 x 3.262) x (km · LY · psc) / (km · LY) x (10¹⁵⁻¹²) =
(0.03388) x (psc) x (10³) =
33.88 parsecs
Length•Width•Height is the answer
Answer:
A sample of 5.2 mg decays to .65 mg or to 1/8 of its original amount.
1/8 = 1/2 * 1/2 * 1/2 or 3 half-lives.
3 * 30.07 = 90 yrs for 5.2 mg to decay to .65 mg
You can get these other numbers similarly:
5.2 / .0102 = 510 requires about 9 half-lives which is 30 * 9 = 270 yrs