Question

In 35.0 s, a pump delivers 550L of oil into barrels on a platform 25 m above the pump intake pipe. The density of oil is .820 g/cm^3. Caculate the work and power done by the pump.

Answer 1

Hint:
the requested work is the work made, by the external force, against the gravity. Do, since both external force (whose, modulus is equal to the weight of the oil) and the displacement traveled by the oil, have the same direction, then we can write:
\[\Large W = {\text{weight }} \times {\text{ distance}}\]
where:
\[\Large {\text{distance}} = 25\;meters\]
and
\[\Large {\text{weight = density }} \times {\text{ gravity }} \times {\text{ volume}}\]
Finally, the requested power P, is given by the subsequent formula:
\[\Large P = \frac{W}{{\Delta t}}\]
where \Delat t is the interval of time, within the external force is acting on the oil, namely:
\[\Large \Delta t = 35\sec \]

Answer 2

I think u need mire help???!!!

Answer 3

*more

Answer 4

U know mass=volume*density

Answer 5

Volume of oil=550L=550dm^3=550*1000cm^3
Right??!!!

Answer 6

Now mass of oil=550*1000*.82gm

Answer 7

Let mass =m

Answer 8

U know work w=mgh=550*1000*981*2500dyne=?

Answer 9

Power p=w/t=?

Answer 10

Given t=35sec