Recent content by JayneDoe

To me, it would not make sense to report an uncertainty with more than one significant figure. Additionally, it wouldn't make sense to report something like '0.377 +/- 0.1' -- because your uncertainty is in the tenths place, the hundredth and thousandth places of the measurement are pretty...

Ah, you're right. Serves me right for not thinking. In that case, since you know the voltage is the same across both resistors, you know that 10 = R_1I_1 = R_2I_2 . If you replace my first equation with this, then I think that the rest of my post is still fine. EDIT: However, now that I think...

Based on the work you've already done, you have the system of equations \begin{cases} 10 = R_1I_1 + R_2I_2 \\ 2.82 = I_1 + I_2 \end{cases} Since I_1 and I_2 are the only unknown quantities, you can easily solve this system for either one. You've stated both that P = \frac{V^2}{R} and V = IR -...

I'm able to get x=0.5kg by that method, actually. If you leave '1/15' on the LHS of the first equation rather than approximating it as '0.06', you should find the same.

It wouldn't really make sense to try to 'put all the constants on the RHS of the equation' - everything in the first equation is a constant and we know that 'x', the only variable in the second equation, cannot be zero. Consider, instead, that 'k', as a property of the spring, will not change...

Starting at the origin does mean that xi=yi=0, yes. I'm not entirely sure what your second question means, but you should be able to formulate an expression for both the x and y position of the particle at any given time and then use substitution to find the answer.