Posted in enart.xn--gmq282eogn.com on the March 20, 2010
An object has a mass of 615 grams and a volume of 105cm^3. What is the
density of the object?And considering that their homework is being being answered for them.Yes, I agree with the person who has answered. This answer can not be made bettermusicfan,
If you were not fully satisfied, you could ask for clarification
before rating my answer. I don't understand why you were not fully
satisfied.
Rainbow~It just boils down to something like this:
Suppose you have a box, with edges each 1 foot long. So it's a box
with a volume of 1 x 1 x 1 = 1^3, or, 1 cubic foot (1 ft^3). Now
suppose you put in some marbles, and we're way out in space,
weightless, so the marbles can all float about inside the closed box
and spread out evenly. If I put in 1 marble, how many marbles per
cubic foot is that? It is, of course, 1 marble per cubic foot, or,
1/ft^3. That's the density of marbles, 1/ft^3.
Now let's put in more marbles. Put 10 marbles into the same box.
What's the density? It's 10/ft^3. It has a density of 10 marbles per
cubic foot.
But each marble has a property called "mass", which effectively is
a measure of how hard it is to change its velocity by pushing on it.
But whatever mass ultimately is, you know that each marble hass mass.
So by having a certain density of marbles per volume of space -a
marble density- you also have a mass density.
Let's say that each marble has a mass of 1 ounce. (Abbreviated 1
oz.) One marble constrained within the box gives the box a mass
density of 1 oz/ft^3. 10 marbles = 10 oz and the volume has a mass
density of 10 oz/ft^3.
But whatever the amount of mass and the amount of volume, to get
mass density all you need do is divide mass (m) by volume (v): m/v.
And it's the same in any system of units. The units in your problem
are grams (mass) and centimeters (distance), with volume being c^3.
Given mass = 615 grams, and volume = 105 cm^3, what then is the mass
density? It is 615/105 ~ 5.86 g/cm^3.#If you have any other info about this subject , Please add it free.# |
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