Leaky sandbag A bag of sand originally weighing 144 lb was lifted at a constant rate. As it rose, sand also leaked out at a constant rate. The sand was half gone by the time the bag had been lifted to 18 ft. How much work was done lifting the sand this far? (Neglect the weight of the bag and lifting equipment.)

Explanation:

Let m be the total mass initially and h be the height up-to which it is lifted.

Rate of fall of sand

= m/(2xh) lb/ft

At any height p , mass leaked out = m/(2xh) x p

At certain height p, mass  will  be m - m/(2xh) x p . This mass is lifted by small distance dp

work done

dW = [m - m/(2xh) x p ] g dp

Integrating it from 0 to h , we get the total value of work done

W = ∫[m - m/(2xh) x p ] g dp

= mg ( p - 1 / 2h x p²/2 )

Now taking limit between 0 to h

W = mg ( h - 1/2h x h²/2)

= mg ( h - h / 4 )

3 / 4 x mgh

= .75 x 144 x 32 x 18

= 62208 ft-poundal .

the diagram which shows forces as either balanced = same speed or unbalanced = accelerarting

a. carbon monoxide from car exhaust

b.particulate matter, such as soot

c.water vapor from a factory smokestack

d.acid rain

these are the answer choices im on that question right now