Question

An object with a mass of 0.255 kg and density of 2.89 g/cm3 measures 34 mm in length and 46 mm in width. What is the height of the object?

A. 5.6 cm
B. 7.2 x 10^-4 cm
C. 5.6 x 10^-2 cm

Answer 1

B. would be 0.00072 and C. would be 0.056 solved. I am just not sure how you would start this?

Answer 2

Volume equals lwh first find the volume, then solve for h.

Answer 3

So like this 34 (l) * 46 (w) * ? (h)

Answer 4

Do I divide the mass 0.255 over the density 2.89?

Answer 5

First find the volume using the density equation.

Answer 6

You will need to convert everything to the same units also.

Answer 7

density = mass / volume right?

Answer 8

right so volume does equal mass/ density

Answer 9

Just remember \(\large \rho = \frac{m}{V}\)

We have \(\large \rho\) and we have \(\large m\) so we just plug in everything we have
*remembering that \(\large V = L\times W\times H\)

So we have

\[\large 2.89\frac{g}{cm^3} = \frac{.255kg}{34mm \times 46mm \times H}\]

Now, make sure everything is in the proper units and then you can solve for H

Answer 10

Note that when I say "Proper units" ...Notice your answer choices are all in the units \(\large cm\)

But here, we have \(\large mm\) and we have \(\large grams\) and we have \(\large kilograms\)

We need to cancel out those grams and kilograms SOMEHOW and we need to turn those mm into cm SOMEHOW

Answer 11

I'll let john take over

Answer 12

so 34 mm equals 3.4 cm and 46 mm = 4.6 cm

Answer 13

Exactly! So that solves our cm problem...now what can we do about the grams and kilograms?

Answer 14

so we have to cancel out the 2.89 grams and .255 kg?

Answer 15

Well not so much "cancel out" but we do need to have the same units on both sides of the equation so the UNITS cancel out

So we can either convert 0.255kg to g or we can convert 2.89g to kg...your choice

Answer 16

so 2.89 grams = 0.00289 kg

Answer 17

That works

So now we have

\[\large 0.00289 \frac{kg}{cm^3} = \frac{0.255kg}{3.4cm \times 4.6cm \times H}\]

And now that we have all the correct units to cancel out...we solve for H

Answer 18

So by cancel out do we have to get rid of the kg's first?

Answer 19

Well, before we get rid of the units...lets make sure they all work....if we solve for H...we multiply both sides by H right? And then divide both sides by \(\large 0.00289 \frac{kg}{cm^3}\) correct?

Answer 20

that sounds right

Answer 21

Great *I'll do it out just so you can see it*

Multiplying both sides by H:
\[\large H \times 0.00289 \frac{kg}{cm^3} = \frac{0.255kg}{3.4cm\times 4.6cm\times \cancel{H}}\cancel{\times H}\]

That gives us

\[\large H \times 0.00289 \frac{kg}{cm^3} = \frac{0.255kg}{3.4cm\times 4.6cm}\]

Then when we divide both sides by \(\large 0.00289 \frac{kg}{cm^3}\) we get

\[\large \frac{H\cancel{\times 0.00289 \frac{kg}{cm^3}}}{\cancel{0.00289 \frac{kg}{cm^3}}}=\frac{0.255kg}{3.4cm\times 4.6cm}\times \frac{1}{0.00289 \frac{kg}{cm^3}}\]

Leaving us with:

\[\large H = \frac{0.255kg}{0.00289\frac{kg}{cm^3}\times 3.4cm \times 4.6cm}\]

Now we can look againat units to make sure

Answer 22

So now, lets just multiply everything on the bottom together...and see what we get *units included*

\[\large H = \frac{0.255kg}{0.0451996\frac{kg\times cm \times cm}{cm^3}}\]

Look right?

Answer 23

I got 5.6 for the 0.255/0.00289*3.4*4.6

Answer 24

Perfect! That's what you should get!

Now we just check the units to make sure it is right

\[\large 5.64\frac{kg}{\frac{kg\times cm \times cm}{cm^3}}\]

We can now see \(\large 5.64\frac{\cancel{kg}}{\frac{\cancel{kg}\times cm^2}{cm^3}}\)

And we can also see

\[\large 5.64\frac{1}{\frac{\cancel{cm^2}}{cm\cancel{^3}}}\]

Leaving us with just \(\large 5.64 cm\) as our answer!

Answer 25

*Or just 5.6 for rounding purposes*

Answer 26

Thank you very much for your time and help :)

Answer 27

and explaining it to me step by step :)

Answer 28

Not a problem! :)