anonymous
  • anonymous
A cylindrical tank with diameter 20 in. is half filled with water. How much will the water level rise if you place a metallic ball with radius of 4 in. in the tank? Give your answer to the nearest tenth. - What causes the water level in the tank to rise? - Which volume formulas should you use?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
pleeeease help
austinL
  • austinL
You have the volume of a cylinder. \[V_{cylinder} = \pi r^2 h\] The volume of a sphere. \[v_{sphere} = \frac{4}{3} \pi r^3\]
anonymous
  • anonymous
So I find the volume of the cylinder and the volume of the sphere and subtract them, right? But, how would I find the volume of the cylinder if I'm not given the height?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

austinL
  • austinL
No, add.
anonymous
  • anonymous
I'm still confused.. what do I do about the height?
austinL
  • austinL
@amistre64
ybarrap
  • ybarrap
The tank will rise according to the new volume in the cylinder: $$ \Large V_{old} = \pi r_{cyl}^2h_{old}\\ \Large V_{new} = \pi r_{cyl}^2h_{old}+ \frac 4 3 \pi r_{ball}^3=\pi r_{cyl}^2h_{new}\\ \Large \implies \pi r_{cyl}^2h_{new} - \pi r_{cyl}^2h_{old}=\frac 4 3 \pi r_{ball}^3\\ \Large \implies h_{new}-h_{old}={\frac 4 3 \pi r_{ball}^3 \over \pi r_{cyl}^2}\\ \Large ={4 ~r_{ball}^3 \over 3 ~r_{cyl}^2} $$ Let me know if you have any questions.
anonymous
  • anonymous
Am I just missing something super simple? What is the height?
amistre64
  • amistre64
how high is "half filled with water" to begin with?
amistre64
  • amistre64
you cannot determine the half volume of the cylindar without knowing how high it is to start with
austinL
  • austinL
That is where I got thrown and called in the cavalry(you @amistre64 )
jdoe0001
  • jdoe0001
heheh
anonymous
  • anonymous
It doesn't say in the book, it just says half filled with water
jdoe0001
  • jdoe0001
@kaileylol can you post a quick screenshot?
amistre64
  • amistre64
|dw:1377639096023:dw|
amistre64
  • amistre64
lets assume the height is suitable to contain the volume of the sphere at least
amistre64
  • amistre64
the height it rises will be the same regardless of its original height
ybarrap
  • ybarrap
They are asking how much the water rises, not what is the new height. So the question is "What is the increase in height" For this, you don't need to know the actual height. The amount of water displaced in cubic inches is equal to the volume of the metallic sphere. This determines the increase in volume. The difference between the new volume and the old allows you to determine the difference in height. This is the increase in the water level.
amistre64
  • amistre64
just determine the volume of the sphere then, and solve for h
amistre64
  • amistre64
\[V_{sphere}=\pi r^2h\] \[\frac{V_{sphere}}{\pi r^2}=h\]
amistre64
  • amistre64
whether its empty, or half full to start with ... the rise in height will be the same with the new volume added in
ybarrap
  • ybarrap
\(\Large V_{old}\) and \(\Large V_{new} \) above are the water level volumes before and after the ball is immersed.
anonymous
  • anonymous
I think my screenshot is a bit late
1 Attachment
ybarrap
  • ybarrap
We are using Archimedes principle: http://en.wikipedia.org/wiki/Boyancy
ybarrap
  • ybarrap
Except, we don't need to worry about density.
jdoe0001
  • jdoe0001
hmm I see
jdoe0001
  • jdoe0001
ybarrap The tank will rise according to the new volume in the cylinder: \(\bf V_{old} = \color{blue}{\pi r_{cyl}^2h_{old}} \implies \textit{ tank without the ball}\\ V_{new} = \pi r_{cyl}^2h_{old}+ \frac 4 3 \pi r_{ball}^3 \implies \textit{ tank with the ball } \implies \color{red}{\pi r_{cyl}^2h_{new}}\\ \textit{subtracting the old tank volume from the new one}\\ \textit{should give the difference, which is the volume of the ball}\\ \large \implies \color{red}{\pi r_{cyl}^2h_{new}} - \color{blue}{\pi r_{cyl}^2h_{old}}=\frac 4 3 \pi r_{ball}^3\\ \pi r^2{cyl}(h_{new}-h_{old})=\frac 4 3 \pi r_{ball}^3\\ \textit{difference between new and old "h" will give the risen amount}\\ \large \implies \color{red}{h_{new}}-\color{blue}{h_{old}}={\frac 4 3 \pi r_{ball}^3 \over \pi r_{cyl}^2}\\ \textit{risen amount}=\Large {4 ~r_{ball}^3 \over 3 ~r_{cyl}^2}\)
ybarrap
  • ybarrap
Thanks for adding that annotation @jdoe0001 ! I had actually added some but my MathJax kept crashing; so I redid without it. I upgraded my Chrome and added a MathJax extension, hopefully this helps.

Looking for something else?

Not the answer you are looking for? Search for more explanations.