Conversion Factors:
Conversion Factors can be written as a fraction, with one side of the formula in the numerator (top) and one side in the denominator (bottom).
Example: ( 1 m / 109nm ) or ( 109nm / 1 m )
Combine the conversion factors in a way, so that meters cancel out, because we don't have meters in the initial problem. Units can cancel, just like any other factor using a fraction.
( 109nm / 1 m ) x ( 1 m = 102cm ) = 109nm / 102cm = 107 nm / cm
This translates to: 1 cm = 107 nm
Convert either unit into the other, using thederivedconversion factor we just came up with. I'm going to convert the 10 nm into cm's because it's easier.
10 nm x ( 1 cm / 107nm ) = 10 / 107 nm = 1 / 106 nm = 10-6 nm
Note: You can move a number raised to an exponent from the numerator to the denominator, or the other way, as long as the polarity of the power changes. Polarity is just negative or positive. This was an optional step, I did it because it just is easier than expressing a fraction.
Now Volume of a Cube, is going to be side cubed: V = s3
Volume of small cubes = (10-6cm)3 = 10-18cm3
Volume of the larger cube = (1 cm)3 = 1 cm3
To figure out how many cubes fit into the larger cubes, divide the volume of the larger cube by the volume of a single smaller cube. The units cancel out, which makes sense because this is a "counting number", that is, you really don't say I have 3 second apples, you just say I have 3 apples.
1 cm3/ 10-18cm3= 1018
Remember, you can move numbers raised to a power along the numerator and denominator as long as the polarity of the power changes.
The answer is 1018 cubes. That's a 1 with 18 zeros following it, orone quintillion.
241920 in^3 == typical minivan cargo volume (V) 17.576 in^3 == volume of a cube surrounding a tennis ball (v) 13764 == V/v = number of tennis ball sized cubes to fit into the cargo space of a minivan
This is a cracking question. [I'm assuming that 'integral side lengths' is intended also to mean integral positions. That is to say, the position vectors of the cubes' vertices should consist only of integer components...such that we are effectively drawing our cube in a 3D grid. If this was not a condition, the answer would be infinity.] So then: first, there are 53 = 125 1x1x1 cubes. Then 43 = 64 2x2x2cubes. 33 = 27 3x3x3 cubes. 23 = 8 4x4x4 cubes. And, stunningly, 13 = 1 5x5x5 cube. So far we have 225 cubes. That would not be interesting. But we have only counted the cubes with edges parallel to the edges of the main cube. Suppose there are some more cubes formed by diagonal lines? Since edge lengths and vertex positions both have to be integral, and we are working in 3 dimensions, we are actually looking for Pythagorean quadruples - integer solutions to a2 + b2 + c2 = d2 Up to now we have only used trivial solutions like this: 52 + 02 + 02 = 52 Our edges have only moved in one dimension. We might consider edges that move in two dimensions, using the smallest Pythagorean triple: 32 + 42 + 02 = 52 But diagonal edges of length five are clearly not going to fit. So introduce the third component, and we find this: 12 + 22 + 22 = 32 By using diagonal edges it is possible to constuct a few more valid 3x3x3 cubes. It turns out, I think, that four of these can be formed. So we have 229 cubes in total.
How big is the marble? If it is 12" across, you'll fit one in there.
from 1 to infinite.. depends how many you want to put in it.
10
There would be 1,452 1cm cubes that fit in a 12cmx12cmx11cm cube. This is determined by multiplying the dimensions of the larger cube together (12x12x11) and dividing by the volume of the smaller cubes, which is 1cm^3.
64
How many cubes fit will depend on the size of each cube.
4 of them can.
5*5*5 = 125 of them.
64 of them
There are: 2400/96 = 25
1 000 000 cubes would be held. 1 000 litres of water.
512
9
1 decimeter = 10 centimeters. So one face of a decimeter cube will have [10 x 10] centimeter cubes, so 100 on each face.
10 along each edge, 36 in all.