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Working out the units?

by tommyboo
Tags: units, working
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tommyboo
#1
Nov8-11, 05:38 PM
P: 10
I am having a problem finding the correct SI unitsfor the quantity A?

In the equation

A=√(R/TY)

That is A equals the square root of R divided by TY

(not to good showing workings on the computer sorry)

, the SI units of the quantity R are kg m^3 s^–2, the SI units of the quantity T are kg and the SI units of the quantity Y are m s^–2. What are the correct SI units for the quantity A?
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Mark44
#2
Nov8-11, 06:29 PM
Mentor
P: 21,397
Quote Quote by tommyboo View Post
I am having a problem finding the correct SI unitsfor the quantity A?

In the equation

A=√R/TY

That is A equals the square root of R divided by TY
Both your notation and explanation are ambiguous.

Is the expression on the right side this?
[tex]\sqrt{\frac{R}{TY}}[/tex]
or this?
[tex]\frac{\sqrt{R}}{TY}[/tex]
Quote Quote by tommyboo View Post

(not to good showing workings on the computer sorry)

, the SI units of the quantity R are kg m^3 s^2, the SI units of the quantity T are kg and the SI units of the quantity Y are m s^2. What are the correct SI units for the quantity A?
tommyboo
#3
Nov8-11, 06:32 PM
P: 10
The first one R/ty all square root. Do apologise for the bad format

gsal
#4
Nov8-11, 07:57 PM
P: 898
Working out the units?

Then, the units of A are meters.
zgozvrm
#5
Nov9-11, 12:47 PM
P: 754
To clarify gsal's answer...

You have the expression
[tex]\sqrt{\frac{R}{TY}}[/tex]

Simply, insert the units for each variable (in place of the variables):
[tex]\sqrt{\frac{\frac{kg\cdot m^3}{s^2}}{(kg)(\frac{m}{s^2})}}[/tex]

and simplify...
[tex]\sqrt{\left(\frac{kg \cdot m^3}{s^2}\right) \left(\frac{s^2}{kg \cdot m}\right)}[/tex]

kg and s2 cancel out, leaving
[tex]\sqrt{\frac{m^3}{m}}[/tex]

which is
[tex]\sqrt{m^2}[/tex]

or, more simply m


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