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**irspow**- Replies: 1

*edit...sorry, I see the error of my thinking...

I was not considering torque only the mass...what I did is simply split the mass in half without considering the torque applied by those masses about the point c...sorry for wasting the space with my oversight....

The center of mass of a cone of uniform density is (0, h/4) relative to the base center, usually stated as 3h/4 from the vertex, according to everybody. Fine. But follow along for kicks...

If we define the right side of the cone by a line we get:

y=h-hx/r

y=(hr-hx)/r

If we integrate for solid of revolution about the y axis x is the radius...

x=(hr-ry)/h and the volume would be given by the integral...

V(y)=p⌠f(y)^2 dy in this case:

V(y)=p⌠((hr)^2+(ry)^2-2hyr^2)/h^2 dy

V(y)=[(pr^2)/h^2]*⌠h^2+y^2-2hy dy, let k=[pr^2/h^2] for simplicity...

V(y)=k[yh^2+y^3/3-hy^2]

V(y)=(k/3)[3yh^2+y^3-3hy^2] Obviously if the interval is y=[0,h] we get the normal Volume equation for a cone...((pr^2/(3h^2)(h^3)=(phr^2)/3))

But if we find a point c where V(y) for y=[h,c] is equal to V(y) for y=[0,c] then we know that the volume above c and below c are equal.

Considering that the cone is assumed to have uniform density, having equal volume above and below c is the same as having equal mass above and below c, the definition of the center of mass...okay so I say let's find c...

If we use the two intervals to find when they are equal and allow the leading constants to cancel out...the two intervals again are [h,c] and [c,0], you get:

3h^3+h^3-3h^3-3ch^2-c^3+3hc^2=3ch^2+c^3-3hc^2

h^3-3ch^2-c^3+3hc^2=3ch^2+c^3-3hc^2

2c^3-6hc^2+6ch^2-h^3=0

if we let h=1, it does not matter, this is just for ease...and we use an interative process such as Newton's approximation...

2c^3-6c^2+6c-1=0

and the approximation being...

(4c^3-6c^2+1)/(6c^2-12c+6) letting c1=0 we see the the sequence...

0, 1/6, 0.2044..., 0.206295.., 0.206299473.., 0.206299474015900262623.., 0.20629947401590026262414718036385, and then next one does not change anything within the next 20 digits...

Why is point c not h/4? Half of the volume, and thus mass, is above c and below c at this point, how is this not the center of mass? (This is assuming uniform density and using sound solid of revolution integration.)

So, I know, and I have seen, that there are proofs out there that c is h/4, but why does integration of a solid of revolution lead to a different result?

Sorry for being dense, I just want to see the error of my logic...thank you for your time.

12*95=1140 if 11 students got 100's the 12th student can have a minimum score of 40

11*100+40=1140 1140/12=95

If you do the long division you get:

2a-1 with the remainder 4

so the answer is 2a-1+4/(a+2)

If L=lighthouse height, C=cliff height, and x=distance from ship to cliff and lighthouse

tan31=(L+C)/x and

tan26=C/x solve for x in this one

x=C/tan26 plug this into first equation

tan31=[(L+C)tan26]/C

Ctan31=Ltan26+Ctan26

Ltan26=C(tan31-tan26)

C=Ltan26/(tan31-tan26)

C=107.78m

So your answer seems good. The above is just how the problem struck me. I just noted the two ratios and used two equations with two unknowns. It seemed pretty straightforward. You did the same thing which worked out fine. The only thing that I did different was to not calculate any values until the very end, because I didn't need them until then to solve for the height of the cliff. (I do this a lot because it keeps small rounding errors from getting very large in problems with many steps.)

Oh, and now we know how far the ship is from the lighthouse too...LOL

x=C/tan26 or more precisely because C was rounded

x=L/(tan31-tan26)

x=220.99m

A variable is just that. It is a symbol to represent any value that is possible within the context it is used as long as mathematical rules are adhered to. You can and often do know the entire range of values possible for a given variable but use it as shorthand for all of those values. Can you imagine listing all the values instead of just x for an equation? In many cases you could not do so. For example:

y=2x perhaps the simplest of terms you will ever come across.

Well, x (and y for that matter) can take on any value. It would be impossible to list every possibility if you had an eternity. But what is powerful about variables is that an infinite number of terms can be expressed by the simple relationship demonstrated by the variables. And more times than not, that is what is really at the heart of all mathematics, logical symbolic relationships among abstract things.

I don't know if any of this helps, but a variable in simplest terms is a symbol that is used to look at mathematical relationships without having to actually assign a specific value at the time you are studying the relationship. It is treated mathematically like any other value, but one need not assign a specific value to it until necessary.

or...

A = s²/(4tan[180/n]) where s is length of side and n is number of sides (for those who don't want to use pi to calculate the area of a polygon)

Well that is more physics than math

Here is the formula you will be using to solve the question:

Ft=p; F=force, t=time, p=momentum or mv (mass times velocity)

* note that we will really be talking about average Force during the collision and the change in momentum over that time period (impulse)

Ft=mv=p

F=p/t

The v (velocity) just before impact and just after impact needs to be found first.

Veloctiy at impact is found by using:

Now to find the velocity just after leaving the ground

Finding delta v and multiplying this to the mass will give us the p we were looking for:

Plugging p into our origianl impluse equation

This is the average force that occurs during contact with the ground.

Hope that helps.

C(f) = (f - 32)(5/9) = (5f - 160)/9;

F(c) = (9c/5) + 32 = (9c + 160)/5;

K(c) = 273.15 + C;

K(f) = (2297.9 + 5f)/9;

Here goes the solutions, I will hide them for those who wish to solve them on their own.

The "bonus" is easier.

As y varies from 1 to ∞ x varies from 1 to 0. But we never integrate in the negative direction, so your translated function now one of f(x) would need to be integrated from 0 to 1.

Okay, I too have no idea how they came up with that function, so uh er, I guess that I can't help you either. Sorry.

**irspow**- Replies: 6

What is the maximum number of 1" spheres that will fit completely within a 24" cube?

As a bonus, what is the percentage of empty volume within this maximized arrangement?

I have found one possible solution although there would be many others along the same lines.

First for the conditions that I used. I arbitrarily chose a distance of 4r to exist between any two glasses' geometric centers. Then I allow the height of the glasses to vary to make this true.

Place two glasses standing up 4r apart. Next, lay a glass on its side touching the two standing ones. Repeat second step again on the opposite side of the standing glasses.

No proof is needed to show that the standing glasses are 4r apart. Also the glasses lying on their sides are 4r apart because they are spaced only by the diameter of the standing glasses and their radii.

Now we let the height vary to accomodate this distance.

Overhead it can be seen that centers along the plane of the table from one lying glass to one standing glass form a triangle two equal sides of 2r. Thus the base of our vertical triangle will be r√8.

The side of our vertical triangle will be (h/2 - r). So the hypotenuse of this triangle will be;

d = distance between geometric centers of lying to standing glass

d = √(8r² + h²/4 + r² - hr);

We want d = 4r;

4r = √(9r² + h²/4 - hr);

4r = (1/2)√(36r² + h² - 4hr);

8r = √(36r² + h² - 4hr);

64r² = 36r² + h² - 4hr;

-h² + 4hr + 28r² = 0;

h = [-4r ± √(16r² + 112 r²)] / -2;

h = [-4r ± √(128r²)] / -2;

h = [-4r ± 4r√8] / -2;

Since h must be positive;

h = (-4r - 4r√8) / -2;

h = [-4r( 1+√8)] / -2;

h = 2r(1 + √8);

So this is just one particular case. If the dimensions of the glasses and the space wanted were known then a solution could be calculated in a similar manner.

Welcome landon! Feel free to stay awhile. It only gets better the more you look around.

If you remember that a cm³ = mL it is more simple.

80g/50ml(1000ml) = 1600g

MathsIsFun, they seem very accessible to me. I could only comment on things to add, but you stated above that you plan on doing such anyway. I would suggest that before you move to more complicated material in algebra, a little chat on order of operations would be critical. We tend to follow those rules without thought at this point in our lives, but for those new to algebra it is paramount to know them.

Order of operations popped into my head while reading the pages because they weren't needed. Anyway, I thought then that it would not be so apparent to a newcomer and also low on our natural list of things to explain. (Because we usually don't really think about it anymore.)