i know that Y is vertical and is voltage as i know current is X and horizontal i meant how is it phrased, as in do i write i will plot the graph of: a) voltage against current (thus y against x) or b) current against voltage (thus x against y) im varying the current hence its on the y axis i just cant remember how to put it in words it is supposed to go in a specific order determined by which ...

2024年6月17日 · This data will be approximately equal to e^x. The bit that is actually varying in this situation is x. And in this situation x is varying linearly. So the solution is to apply the natural log to your results so that you can get that value. And because that value varies in a linear fashion your graph will be a straight line.

2020年4月2日 · I think that the only easy way to do it is to plot both log(y) against x (in which case use y=ab x) and log(y) against log(x) (in which case use y=ax n). But it is very unlikely you will have to do this yourself, simply because of the time and effort involved. That's why they've tipped you off about the relationship to use.

2024年5月16日 · Yes, by convention, "a" would be on the y-axis and "b" on the x-axis. The same would apply if you were asked to plot "a" vs "b". Alternatively, you could remember that the independent variable goes on the x-axis and the dependent variable on the y-axis.

2024年6月3日 · I'm not sure what you mean, but if you have the graph of y = a x n y = ax^n y = a x n, you can take logs of both sides to give: log ⁡ (y) = log ⁡ (a x n) \log (y) = \log (ax^n) lo g (y) = lo g (a x n) Using log rules, this is the same as: log ⁡ (y) = n log ⁡ (x) + log ⁡ (a) \log (y) = n \log (x) + \log (a) lo g (y) = n lo g (x) + lo g ...

2024年6月17日 · When you plot the graph of Resistance against Length. The gradient of graph will give the value for (resistivity / cross-sectional area) If you equate the worked out gradient to that equation you will get gradient = resistivity / cross-sectional area re-arrange to get gradient * cross-sectional area = resistivity EDIT: If you compare the equation for …

2024年6月4日 · Hi, The question is as follows: (Two quantities Q and H are believed to be related by the equation Q=kH^(n). The experimental values obtained for Q and H are shown below: Q = 0.16 0.20 0.27 0.34 0.40 0.47 0.55 H = 1.14 1.78 3.24 5.14 7.11 9.82 13.44 Determine the law connecting Q and H, showing on your graph the point of interception and the gradient.)

There are a few clues you could use-if one of them is time then that would go on the x axis. If you're given A in terms of B then that would probably indicate that A should go on the y axis. Otherwise, just take your pick but obviously make sure you label the axes.

2024年5月24日 · F=GMm/(r^2) Rewriting this a little u get F=GMm*(1/r^2) Comparing this to y=mx c u get he gradient as GMm, the intercept is the origin and the x axis must be 1/r^2. For a graph as in c, consider R^2=0.5, F=2GMm R^2=1, F=GMm R^2=2, F=GMm/2 As r^2 approaches zero F approaches infinity therefore the graph is not linear.

2024年6月14日 · Graphing something against something is always Y against X. Also, time tends to go along the x axis as a rule (not 100% of the time though.) Apart from Iodine/Bromine clock experiments (did iodine clock today, got them on the brain lol)