I have the question "An elastic cord of unstretched length $160$ mm has a cross-sectional area of $0.64$ mm$^2$. The cord is stretched to a length of $190$ mm. Assume that Hooke's Law is obeyed for this range and that the cross-sectional area remains constant. The Young's modulus for the material of the cord is $2.0 \times 10^7$ Pa. Calculate the tension in the cord at this extension."
So I know that the Length is given by $L = 0.16$ meters,
Area $A = 6.4 \times 10^{-4}$ square meters,
Extension = $0.03$ meters,
Young's modulus = $2.0 \times 10^7$ Pa.
We have to use the equation:
$$E=\frac{F\times L}{A \times Extension}$$
And rearrange to make $F$ (tension) the subject.
I am not sure how to rearrange this to make $F$ the subject. Also, are the current steps that I am taking correct ? If not could you point out where I am going wrong ? I have also converted the units from mm to m.
The final answer should be $2.4$ Newtons (N).
The area is $0.64\, mm^2=6.4\times 10^{-7}\,m^2$
Then $F=\dfrac{E·A·Extension}{L}=\dfrac{2.0\times10^7\,Pa·0.03\,m·6.4\times10^{-7}\,m^2}{0.16\,m}=2.4\,N$