Not sure how to solve Pascals Law

How is the following problem solved:

ΔP = ρgΔh

400 kPa = (9.8 m/s^2)(1000 kg/m^3)(Δh)

Δh=40.8

I understand that 40.8 comes from dividing 400 by 9.8 but what happens to the 1000? Do you just ignore it because it is a constant?

#Mechanics

Easy Math Editor

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Comments

Notice the units carefully. The pressure is expressed as kPa. Note that 1 kPa = 1000 Pa. So 400 kPa = 400000 Pa. So your calculation effectively becomes:

$400000 Pa = (9.8 m/s^2)(1000kg/m^3)(\Delta h)$

Which leads to the result of $\Delta h = 40.8 m$

Karan Chatrath - 1 year, 11 months ago

Thank you for your answer. This may sound dense but how did you know to do that? If you converted kilopascal to pascals why didn't you convert kilograms to grams?

Taylor Q - 1 year, 11 months ago

Good question. This is because the unit grams is not the SI unit for mass. The unit kilograms is the SI unit for mass. All units for calculation need to be consistent for each quantity. Otherwise, answers become meaningless. SI units comprise of: meters for length, kilograms for mass, seconds for time. Pa is the SI unit of pressure which can be derived from the fundamental SI units of length, mass and time.

I strongly recommend that you read a little bit about the SI and CGS system of units. I also suggest you read some notes on dimensional analysis of physical quantities.

Karan Chatrath - 1 year, 11 months ago

@Karan Chatrath – I will do that. Thank you for the quick thorough response, it makes much more sense now.

Taylor Q - 1 year, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$