A compound is made up of two or more of what?

Physics

2 answers:

dlinn [17]3 years ago

7 0

A compound is made up of two or more elements~
Hope this helps~

jolli1 [7]3 years ago

5 0

A compound is made up of two or more of atoms that are bonded together with each other via a chemical bonding. It is possible to break the compounds into separate smaller components by the process of chemical reaction. One very simple example of compound is water as it is made of two atoms of hydrogen and one atom of oxygen. Whenever atoms are joined together to create a compound, the atoms definitely loose their individual atomic properties. Other very commonly found compound is carbon dioxide. It is made up of one atom of carbon and two atoms of oxygen.

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You're driving in a car at 50 km/h and bump into a car ahead traveling at 48 km/h in the same direction. the speed of impact is

salantis [7]

To solve this problem, we must remember about the law of conservation of momentum. The initial momentum mist be equal to the final momentum, that is:

m1 v1 + m2 v2 = (m1 + m2) v’

where v’ is the speed of impact

Since we are not given the masses of each car m1 and m2, so let us assume that they are equal, such that:

m1 = m2 = m

Which makes the equation:

m v1 + m v2 = (2 m) v’

Cancelling m and substituting the v values:

50 + 48 = 2 v’

2 v’ = 98

v ‘ = 49 km/h

<span>The speed of impact is 49 km/h.</span>

6 0

2 years ago

#1 Not sure where to start. This is for AP Physics!

yaroslaw [1]

First,

$\rho=\dfrac mV$

where $\rho$ is density, $m$ is mass, and $V$ is volume. We can compute the volume of the roll:

$2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V$

$\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3$

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness $x$ . Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

So we have

$V=Ax$

where $A$ is the given area, so

$4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x$

$\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}$

If we're taking significant digits into account, the volume we found would have been $V=470\,\mathrm m^3$ , in turn making the thickness $x=250\,\mathrm{mm}$ .