# Working out the units?

by tommyboo
Tags: units, working
 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?
Mentor
P: 21,397
 Quote by tommyboo 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?
$$\sqrt{\frac{R}{TY}}$$
or this?
$$\frac{\sqrt{R}}{TY}$$
 Quote by tommyboo (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?
 P: 10 The first one R/ty all square root. Do apologise for the bad format
 P: 898 Working out the units? Then, the units of A are meters.
 P: 754 To clarify gsal's answer... You have the expression $$\sqrt{\frac{R}{TY}}$$ Simply, insert the units for each variable (in place of the variables): $$\sqrt{\frac{\frac{kg\cdot m^3}{s^2}}{(kg)(\frac{m}{s^2})}}$$ and simplify... $$\sqrt{\left(\frac{kg \cdot m^3}{s^2}\right) \left(\frac{s^2}{kg \cdot m}\right)}$$ kg and s2 cancel out, leaving $$\sqrt{\frac{m^3}{m}}$$ which is $$\sqrt{m^2}$$ or, more simply m

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