Wiki User
∙ 12y ago27
Wiki User
∙ 12y agoTo find out how many gold bars can fit in the vault, calculate the volume of one gold bar: 6 cm x 2 cm x 3 cm = 36 cm3. Then divide the vault's volume of 500 cm3 by the volume of one gold bar, which is 36 cm3. Therefore, you can fit approximately 13 whole gold bars in a vault with a volume of 500 cm3.
You can find molar volume by dividing the volume of a gas by the number of moles of gas present. The equation to calculate molar volume is V = nRT/P, where V is volume, n is the number of moles, R is the ideal gas constant, T is temperature, and P is pressure.
The correct phrase is "wholesale changes," which refers to making comprehensive or extensive modifications or adjustments to something. "Whole sail changes" is not a common expression in English.
Early chemists discovered that when gases react with one another, they do so in simple whole-number ratios by volume, known as the law of combining volumes. This led to the development of the law of multiple proportions, which states that when elements can combine to form different compounds, the masses of one element that combine with a fixed mass of the other element are in ratios of small whole numbers.
The clear liquid portion of blood, known as plasma, makes up about 50% of blood volume. Plasma contains water, electrolytes, proteins, hormones, and waste products, and plays a crucial role in transporting nutrients and waste throughout the body.
THAT IS PRIVATE INFORMATION. Everyone's family whole names are different. I will not share that information.
First you must find the volume of the gold bar. Volume = Length * width * height V = 6*2*3 V = 36cm3 Then you can divide the vault's volume by the gold bar's volume to find out how many can fit. 500/36 = 13.888888... Roughly, 14 bars could fit into the vault.
Vault only has one syllable. This means the whole word is the syllable.
No, you need to accidently the whole purifier
The question is underspecified. You need know the dimensions of the vault to calculate how many rigid cuboids will fit. For example, the vault could be 0.1 cm wide, 0.1 cm high and 50,000 cm long - giving it a volume of 500 cubic cm. Yet not one single bar will fit into it because the vault is too narrow. The volume of a single cuboid is length * height * width = 6*2*3 = 36 cubic cm. In principle, the number of such bars that can be fitted into the vault is (500/36) - rounded down. That is, 13 bars. In practise, the shape of the vault may reduce this number.
This is an integer - a whole number.I suppose you could divide it by 1 . . .
You select value of n so that n-15 is a whole number.
Hematocrit.
its different because a volume is how much and area is the whole thing
When I tried it; we were blasted the whole night. We smoked the whole gram out of a bowl and Bong. But, you're suppose to hit it once off the bowl and hold it in for 15 seconds. That's suppose to leave a 15 minute high.
Well, gymnastics is a sport itself with a whole bunch of events in it. For example: the women have vault, uneven bars, beam, and the floor exercise. The men have the pommel horse, the rings, the parallel bars, the "mushroom", floor, vault, and a single bar.
They where home schooled there whole life jkjk how am I suppose to know
I suppose you could, but it will probably come with a whole host of stds.