Sorry for not specifying this initially. The top row is the values on the x-axis: the period T in seconds. The bottom row is the values on the y-axis: \sqrt{m} in g. The masses were 100 g, 200 g, 300 g, 400 g, and 500g.
After tons upon tons upon tons of Googling, I determined that the spring constant is correct if the input mass is given in kg, not g. Dividing by 1000 gives roughly 167, which seems much more correct. Woohoo!
Not woohoo. 167 is still incredibly high for the spring constant. The ultimate goal...
Homework Statement
Construct a plot of T versus \sqrt{m}. Fit a straight line through your data and use Eq. (3.2.3) to obtain the spring constant k from the slope, as well as an uncertainty estimate for k.
T is the period of an inertial balance with various masses of mass m. Periods are...
Ah, I misremembered (if that's a word) how to do it. I've think I've got it now!
u = 500 - 0.4r
du = -0.4r dr
-2.5 du = 1 dr
\int \frac{-2.5}{u}du
-2.5 ln (u) + C = t ( = t pulled from earlier)
-2.5 ln(500 - 0.4r) = t
Then I have to solve that for r, which gives me r =...
Okay, so I think I remember now. If u = 500 - 0.4r, then the following would be my work.
u=500-0.4r
du=dr
\int \frac{1}{u}du = ln (u) = ln (500 - 0.4r)
This isn't correct though, at least according to Wolfram|Alpha.
I remember the right side, yes, it's the u-substitution that I can't remember. I've been looking some stuff up about it to try to jog my memory; we'll see if I get anywhere.
Okay, that makes sense. So the final equation should therefore be \frac{dr}{dt} = 500 - 0.4r? It's been 3 years since I took calc I, and I just discovered that I don't remember how to do integrals. I have the equation up to \int \frac{1}{500-0.4r}dr =\int dt, but I can't get any further (hence...
If r' (or \frac{dr}{dt}) is the rate of change of the drug in the blood stream, wouldn't r then be the total amount currently in the blood stream? In response to your question, if y' is the rate at which the drug is entering the blood stream, the total amount would be y - (rate at which it is...
The rate in is constant at 500 mg/cm3/h, but the rate out varies based on the total amount in the bloodstream. So, in the differential equation, should it therefore be 500 instead of 500t? I think the \frac{-0.4}{t} is still correct. Is that the case? My formula would, therefore, become...
Homework Statement
A certain drug is being administered intravenously to a hospital patient. Fluid containing
5 mg/cm3 of the drug enters the patient’s bloodstream at a rate of 100 cm3/h. The drug
is absorbed by body tissues or otherwise leaves the bloodstream at a rate proportional to
the...
Okay, I apologize for being an idiot. I was looking at the t > 0 in the problem and thinking that's what I was solving for.
The corrected final equation should be t^{r}(r^{2}-14r+48) = 0, as Mark44 said. Using that formula, r = 6 and r = 8. Submitted those and they are, in fact, correct...
Ah, okay, for some reason I canceled it out rather than writing t^r. Copy-pasting my work from above and editing appropriately now gives me the below output, which should be correct.
(t^2)(r-1)(r)(t^{r-2}) - (13)(t)(r)(t^{r-1}) + (48)(t^r) = 0
(t^r)(r-1)(r) - (13)(r)(t^r) + (48)(t^r) = 0...
I'm trying to simplify it (that's what's getting me), and I can simplify it to (t^r)(r-1) - 13 + 48. I'm clearly doing something wrong, but I don't know what.
It is 48, my apologies. Thanks!