## Homework Statement

I've an equation here: h = 3270 Pa / 1,0 g/cm^3 × 9,81 m/s^2 = 33 cm

How did those units cancel out to make it cm? That's something I don't get.

Given above.

## The Attempt at a Solution

I've tried to cancel them out by looking at the same unit in the numerator and denominator such as, for example m / m , so m cancels out and so on but I don't get it. Does anyone know or can anyone explain why the answer is in centimeters? How do the units cancel out in the given equation? Thank you

Orodruin
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Forum rules require you to show your attempt so you need to actually show your attempt at cancelling out the units.

Forum rules require you to show your attempt so you need to actually show your attempt at cancelling out the units.

If we count it:

h = 3270 Pa / 1,0 g/cm^3 × 9,81 m/s^2 = 333 m

It should be 333 meters. How that becomes centimeters? As for your statement, there really isn't much to show. I can't / don't know how to cancel them out but above I explained the idea of canceling out, so I've proved that I know it.

Orodruin
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If we count it:

h = 3270 Pa / 1,0 g/cm^3 × 9,81 m/s^2 = 333 m

It should be 333 meters. How that becomes centimeters? As for your statement, there really isn't much to show. I can't / don't know how to cancel them out but above I explained the idea of canceling out, so I've proved that I know it.
You cannot put in different units and just assume that the answer will be in SI units. If you did the cancellation correctly, there would be a lot to show. Note that your density is in g/cm^3, not in kg/m^3!

As a hint: How many grams are there in 1 kg? What is (1 kg)/(1 g)?

You cannot put in different units and just assume that the answer will be in SI units. If you did the cancellation correctly, there would be a lot to show. Note that your density is in g/cm^3, not in kg/m^3!

As a hint: How many grams are there in 1 kg? What is (1 kg)/(1 g)?

There are 1000 grams in 1 kg. I still don't get it.

Orodruin
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There are 1000 grams in 1 kg. I still don't get it.
So how many kg/m^3 are there in 1 g/cm^3?

So how many kg/m^3 are there in 1 g/cm^3?

Well, 1 kg/m^3 = 0,001 g/cm^3 --> did you mean it vice versa? If so, then 1 g/cm^3 is 1000 kg/m^3

Orodruin
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Well, 1 kg/m^3 = 0,001 g/cm^3
No, this is wrong. 1 m^3 is not the same as 1 cm^3.

No, this is wrong. 1 m^3 is not the same as 1 cm^3.

That's right but 1 m^3 can be converted into cm^3, for example. I'd just like to let you know that I still haven't got the idea.

Orodruin
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That's right but 1 m^3 can be converted into cm^3, for example.
Yes it can, just like 1 g can be converted into kilograms, the point is that you didn't do that. Hence your answer actually has the units cm^3 kg / (m^2 g).

Yes it can, just like 1 g can be converted into kilograms, the point is that you didn't do that. Hence your answer actually has the units cm^3 kg / (m^2 g).

How can I solve this problem?

Orodruin
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How did you solve the problem of figuring out how many grams are in a kilogram? How would you solve the problem of figuring out what 30 km/h would be in m/s?

How did you solve the problem of figuring out how many grams are in a kilogram? How would you solve the problem of figuring out what 30 km/h would be in m/s?

1 kg = 1000 g

kg hg dag g , so
1 0 0 0

30 km/h can be converted into m/s by dividing it by 3,6 , so it'd be 8,333333... m/s

However, I don't still get the idea in this specific instance.

Orodruin
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You are missing the point entirely.
30 km/h can be converted into m/s by dividing it by 3,6
The point was not what is the actual factor, the point is how you obtain that factor.

You are missing the point entirely.

The point was not what is the actual factor, the point is how you obtain that factor.

Well 1 km = 1000 meters and 1 hour is 3600 seconds. That gives me the factor.

Orodruin
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Well 1 km = 1000 meters and 1 hour is 3600 seconds. That gives me the factor.
So what is cm^3 kg / (m^2 g) in cm? The procedure is exactly the same.

So what is cm^3 kg / (m^2 g) in cm? The procedure is exactly the same.

I've got no idea.

Orodruin
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Yes you do, you did it when you converted km/h to m/s.

So what is cm^3 kg / (m^2 g) in cm? The procedure is exactly the same.

cm^2 * kg / m^2 * g

m^2 / m = m

kg / g = 1000 g => cm * 1000 ?

Orodruin
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m^2 / m = m
Where did this come from? The length units you have are cm^3/m^2 = cm (cm/m)^2.

Where did this come from? The length units you have are cm^3/m^2 = cm (cm/m)^2.

I don't know how to do it, so could you possibly show me?

jbriggs444
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I don't know how to do it, so could you possibly show me?
The classical way to do it is algebraicly, multiplying by a series of "1"s that are chosen so that the units cancel.
For instance, you can write 1 as ##\frac{1\ hr}{3600\ sec}##
$$30 \frac{km}{hr} \cdot \frac{1\ hr}{3600\ sec} = \frac{1}{120} \frac{km}{sec}$$
$$\frac{1}{120} \frac{km}{sec} \cdot \frac{1000\ m}{1\ km} = \frac{1000}{120} \frac{m}{sec}$$

jack action
robphy
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The classical way to do it is algebraicly, multiplying by a series of "1"s that are chosen so that the units cancel.

I call that the Cleverly-Multiply-by-1 method.

I prefer (and strongly advocate to my students) the Substitution method.
So, we have
$$30 \frac{\rm\ km}{\rm\ hr}=30\frac{\rm\ km}{(60\rm\ min)}=30\frac{\rm\ km}{(60\rm (60\rm \sec))}=\frac{30}{(60)(60)} \frac{\rm\ km}{\rm\ sec}$$
$$\frac{1}{120} \frac{\rm\ km}{\rm\ sec}=\frac{1}{120} \frac{\rm\ (1000\ m)}{\rm\ sec} = \frac{1000}{120} \frac{\rm\ m}{\rm\ sec}$$
and the classics
$$1\rm\ m^3=1\rm (100\rm\ cm)^3=10^6\rm\ cm^3$$
$$1\rm\ cm^3=1\rm (10^{-2}\rm\ m)^3=10^{-6}\rm\ m^3$$
and
$$(1\rm\ y)=(1\rm\ (365\ days)) =(1\rm\ (365\ (24\ h))) =(1\rm\ (365\ (24 (60\ m)))) =(1\rm\ (365\ (24 (60 (60\ s)))) =(365)(24)(60)(60)\ s$$

vela and jbriggs444
Mister T
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If so, then 1 g/cm^3 is 1000 kg/m^3

Right. Use the latter instead of the former in your original equation and you will then see how to get the 33 cm answer.

Note that in that original equation it's not just the units that don't work out right, it's also the numbers! I'm not sure where that original equation came from, but whoever did it needs to understand that you can't just mix units together like that and expect to get an answer that has any meaning.

Orodruin
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I don't know how to do it, so could you possibly show me?
If you want spoon fed answers you are in the wrong place. We encourage students to think for themselves and offer guidance. Saying ”I have no idea, show me” is counter productive and will not help you in the long run.

If you want spoon fed answers you are in the wrong place. We encourage students to think for themselves and offer guidance. Saying ”I have no idea, show me” is counter productive and will not help you in the long run.

I have tried to think about it myself as you have probably noticed. Saying that I don't get the idea and asking for help should be completely fine. However, I thank you for giving me this tip.

Orodruin
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I have tried to think about it myself as you have probably noticed. Saying that I don't get the idea and asking for help should be completely fine. However, I thank you for giving me this tip.
If you have really tried to think about it you have unfortunately done a bad job in communicating it. Your posts have been very brief with no argumentation. To show that you have thought about it, you need to describe what you are doing and why you are doing it. This is the only way that we can follow and correct your thought process.

I am not saying this to be mean or to belittle your work. I am trying to give you an honest advice on how to benefit more from and getting the appropriate guidance.

Mister T
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I have tried to think about it myself as you have probably noticed.

Yes, but explaining your thinking is the only path to success. If all you can do is try, and you are not able to explain what you've tried, you will never be able to interact with others in a way that advances your knowledge and promotes your success.

By the way, did you read Post #24?

jbriggs444
Orodruin
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Note that in that original equation it's not just the units that don't work out right, it's also the numbers! I'm not sure where that original equation came from, but whoever did it needs to understand that you can't just mix units together like that and expect to get an answer that has any meaning.
I think the numbers work out quite well to the given answer:
https://www.wolframalpha.com/input/?i=(3270+Pa)+/+(1.0+g/cm^3+*+9.81+m/s^2)
The natural interpretation is the height difference required for a pressure difference of 3270 Pa in a medium with density 1 g/cm^3 and a gravitational field of 9.81 m/s^2. The pressure differential given is about 0.03 bar and the density that of water. This result makes perfect sense and is compatible with the fact that (as any scuba diver knows) the pressure at a depth of 10 m is roughly 2 bar (1 bar from the atmosphere and 1 bar from the 10 m water column, meaning 0.1 bar/m).