Jump to content

Cube and Spheres


Commander

Recommended Posts

Fixed version:

Spoiler

Volume of cube is

[math]x^3[/math]

The volume of a sphere inscribed in a cube is:

[math]\frac{4}{3} \pi \frac{x}{2}^3[/math]

The volume of a sphere superscribed is:

[math]\frac{4}{3} \pi \sqrt{ 3 \frac{x}{2}^2}^3[/math]

4 hours ago, Commander said:

(1) What is the fraction of the volume of *C* is contained in *S1*

[math]\frac{\frac{4}{3} \pi (\frac{x}{2})^3}{x^3}=\frac{\pi}{6}= 52.4\%[/math]

  

4 hours ago, Commander said:

(2) How many times the volume of *S2* is that of *S1*

 

[math]\frac{ \frac{4}{3} \pi \sqrt{ 3 \frac{x}{2}^2}^3 }{\frac{4}{3} \pi \frac{x}{2}^3}=5.2[/math]

 

Edited by Sensei
Link to comment
Share on other sites

8 minutes ago, Genady said:

How much water is needed to exactly cover the balls compared to the volume occupied by the balls?

Spoiler

The answer depends on the density of the balls.. 😛

If their density is less than 1g/mL of water, they will float on the surface and will only be partially submerged.

 

Link to comment
Share on other sites

4 minutes ago, Sensei said:
  Reveal hidden contents

The answer depends on the density of the balls.. 😛

If their density is less than 1g/mL of water, they will float on the surface and will only be partially submerged.

 

Spoiler

Correct. The answer also depends on their chemical composition.

 

Link to comment
Share on other sites

Spoiler

let x=cylinder radius

and x/2 is ball radius

So cylinder v at given height = hπx2

then v ball is 4/3π(x/2)3

h=x/2 + x/2 + x/2(21/2

(simple Pythagorean geometry)

So then plug in a number for x, like 10.

Then h = 5+5+5(21/2)=17.07, ergo undisplaced volume is roughly

5363.

balls volume is

4 x (4/3π53) = 4x523.6=2094

Then, take 5363  - 2094 = 3269

So....cylinder has 3269 units of submerging liquid with balls volume of 2094. (assuming balls denser than liquid)

Roughly 1.56.  (some rounding errors may have accrued)

 

 

 

 

 

Link to comment
Share on other sites

8 minutes ago, TheVat said:
  Hide contents

let x=cylinder radius

and x/2 is ball radius

So cylinder v at given height = hπx2

then v ball is 4/3π(x/2)3

h=x/2 + x/2 + x/2(21/2

(simple Pythagorean geometry)

So then plug in a number for x, like 10.

Then h = 5+5+5(21/2)=17.07, ergo undisplaced volume is roughly

5363.

balls volume is

4 x (4/3π53) = 4x523.6=2094

Then, take 5363  - 2094 = 3269

So....cylinder has 3269 units of submerging liquid with balls volume of 2094. (assuming balls denser than liquid)

Roughly 1.56.  (some rounding errors may have accrued)

 

 

 

 

 

Yes! +1

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.