Physics, 02.09.2019 23:10 randenbrown1653

For this example, let
$m_a and m_b$ are masses
l₁, l₂, l₃ and x be lengths
t₁ and t₂ be times
a mass density is defined as mass per volume, and a volume has a dimension of [l]³. which of the following could be mass densities?
a. $\frac{m_a. m_b}{l_1.l_2}$
b. $\frac{m_a}{l_1^2.l_2^2}.x[/tex ]br / c. [tex]\frac{l_1.l_2.l_3}{m_a}$ < br /> d. $\frac{m_a}{l_1.l_2.\frac{x}{t_ 1}t_2}$

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Answer from: chrisraptorofficial

The answer can be b $\frac{m_{a} }{{l_{1}}^{2}.{l_{2}}^{2}}.x$ i

or d. $\frac{m_{a} }{l_{1}.l_{2}.\frac{x}{t_{1}}.t_{2}}$

Explanation:

As the question says the mass density has units of mass per volume, that means that when you have numbers, this will have units so when you realice the all the operations of the equation the units that the number must have should be $\frac{Mass}{Lenght^{3} }$ .

To get to this result you have to take into account the properties of exponents.

- If you have the same base, and the varibles are multiplying you have to add the exponents for example:

mass x mass each mass has exponent one so you add the exponents (1+1=2) at the end you will have $mass^{2}$ .

- If you have the same base, and the variables are dividying you have to substract the exponents for example:

$\frac{time}{time}$ , each time has exponent one so if you substract (1-1=0) at the end you will have $time^{0}$ . In this case there is an other property that say that when you have any number with exponent cero that number becomes one so $time^{0}$ = 1.

- If we have a negative exponent it means that it must be in the lower part (divisor) of the division for example:

$x^{-2}= \frac{1}{x^{2}}$

Now coming back to the problem we are going to realice each one of the process of the equations to determine which of it fulfills the requirements for mass density remember at end we should get $\frac{Mass}{Lenght^{3} }$ units.

- a) $\frac{m_{a}.m_{b}}{l_{1} .l_{2}}$

In this answer we multiply ma x mb so at the end the units of the upper(dividend) part must be $m^{2}$

In the lower part (divisor) we must multiply l1xl2 so we will have units $l^{2}$

So at the end the units for a option are: $\frac{m^{2} }{L^{2}}$ which doesn't correspond with the mass density units.

- b) $\frac{m_{a} }{{l_{1}}^{2}.{l_{2}}^{2}}.x$

First in the dividend we must multiply ma with x so the units will be m.L because they dont have the same base so we can´t add them

After that we will multiply in the divisor ${l_{1}}^2.{l_{2}}^2$ so as the exponents are 2 we will have 2+2 this is 4. So we will have $L^{4}$

At the end we will have the following units: $\frac{m.L}{L^{4}}$ we have L in both the uper and lower part as we have the same base we can substract 1-4 = -3 as it is a negative number it must be in the divisor so at the end we will have the followind expresion:

$\frac{m}{L^{3}}$ that has the units of mass density

- c) $\frac{l_{1}.l_{2}.l_{3}}{m_{a}}$

First in the dividend we have to multiply l1xl2xl3 so for calculating the units we have to add all the exponents because they are multiplying and have the same base so l1xl2xl3 = $l^{1+1+1}=l^{3}$

The units in the divisor remain in the same way because they doesn't have any operation so we will have mass units s

At the end the units of this expresion will be: $\frac{l^{3} }{m}$ it is the oposite to what we need to have mass density units so this is not the correct answer either

In this moment we have already discard the other options so we know the answer is d but here it comes the explanation

- d) $\frac{m_{a} }{l_{1}.l_{2}.\frac{x}{t_{1}}.t_{2}}$

In this case we have in the dividend mass units so we leave it like that

For the divisor we are going to take all the units and multiply them so remember that l1, l2, l3 and x have length units that we will represent with an L and t1 and t2 have time units that we will represent with a t. So if we replace with the units in the dividend we will have the following expresion }(remeber in this moment we are just going to work with the lower part of the expresion):

$l_{1} .l_{2} .\frac{x}{t_{1} } .t_{2} = L.L.\frac{L}{t}.t\\$

As you can see we have 3 lenghts that are multyplying so we have the same base we have to add the exponents = $L.L.L = L^{1+1+1} = L^{3}$

But we still have time so the expresion in this point is $\frac{L^{3}.t }{t}$ we have two times that are dividying so the operation will be

$\frac{t}{t} = t^{1-1} =t^{0}=1$

At the end we will have: $L^{3}.1 = L^{3}$

Now remember that we were working just with the divisor so as we take into account the mass unit of the dividend the expresion will be:

$\frac{m}{L^{3}}$

So you have with this expression mass density units

Answer from: Quest

the answer is b i think

Answer from: Quest

spectral lines lyman is one set

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