G^−m ÷ g^n
1st g^−m=1/g^m, hence g^−m ÷ g^n = (1/g^m) /(g^n)==> 1/(g^m)(g^(n)
==> 1/(g^m+n) or g^(-m-n)
Answer:
<em>908360.67 lb-ft</em>
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Step-by-step explanation:
height of tank= 6 ft
diameter of the tank = 4 ft
density of water p = 62.4 lbs/ft
A is the cross sectional area of the tank
A = ![\pi r^{2}](https://tex.z-dn.net/?f=%5Cpi%20r%5E%7B2%7D)
where r = diameter/2 = 4/2 = 2 ft
A = 3.142 x
= 12.568 ft^2
work done = force x distance through which force is moved
work = F x d
Force due to the water = pgAh
where g = acceleration due to gravity = 32.174 ft/s^2
Force = 62.4 x 32.174 x 12.568 x 6 = 151393.44 lb
work done = force x distance moved
work = 151393.44 x 6 = <em>908360.67 lb-ft</em>
Answer:
Ana is 12
Step-by-step explanation:
5x4=20
20-8=12