Question

Find the amount of work done in emptying a cylindrical tank completely filled with kerosene. Assume that the radius of the cylindrical tank is 3 m, height is 6 m, and the density of kerosene is 817.15 kg/m3.

Answer 1

slice it up into circular layers of thick ness dh. then work to lift a slice is (3-h).9.81.\[\Pi\]r^2.dh. integrate this.

Answer 2

depends on how you empty it

Answer 3

right. i'm assuming you siphon out water from the top. leaking the tank takes no effortr ofc.

Answer 4

id consider it as how much it weighs with respect to how far you move it;

and the weight is a product of volume

Answer 5

the volume of any given part is: 3^2 pi * \(\triangle h\)

817.15(9 pi)\(\triangle h\) is how much it will weigh as long as i dint forget to convert any measurements

3 m, height is 6 m, and the density of kerosene is 817.15 kg/m3.

Answer 6

the distance we move it, assuming we siphon out the top, changes 0 to 6, or rather 6-h for any given protion

Answer 7

Your values look good to me.
You have \[\frac{g}{m^3}*m^2*m\]
Which leaves you with grams.

Answer 8

Rather kgs :)

Answer 9

\[\lim_{\triangle h->0}\sum_{h=0}^{6} 817.15(9pi)(6-h)\triangle h \implies\int_{0}^{6}817.15(9pi)(6-h) \ dh\]

Answer 10

EXPAND AND INTEGRATE ID ASSUME :)

Answer 11

..... MY LITTLE PINKY FINGER HATES ME :)

Answer 12

Better than having the grey matter hate you ;)

Answer 13

i get abt 415,879 if i did it right

Answer 14

since the containers shape stays constant; this pretty much produces a line to find the area under on the graph

Answer 15

lmao. These are the options to choose from amistre
5.22 x 106 J
4.08 x 106 J
1.69 x 106 J
0.54 x 106 J

Answer 16

for the 106 i meant 10^6

Answer 17

i wonder what they used for an approximation ....

Answer 18

Lol. I wish they would tell me

Answer 19

http://www.wolframalpha.com/input/?i=convert+415668kg-meters+to+joules

4.076 is what this converts to

Answer 20

Where is the work done by gravity come into play?