Recent content by Theorγ

Whops my bad, Vickson is right about removing the units, although I only placed them there initially so you would know where those numbers came from. The solution of the left integral would just actually come from partial fractions method though.

Why neither? And by the way, I'm asking this question primarily for integrating and finding an equation for the velocity of an object at a certain time given that gravity and drag are affecting it. I thought the signs of those respective forces were somewhat relevant.

I assume you are able to solve separable differential equations right?

Set up the sum of forces: \Sigma F = F_{g} - F_{A} \Sigma F = (100 kg)(10 \frac{m}{s^{2}}) - 0.3v^2 ma= (100 kg)(10 \frac{m}{s^{2}}) - 0.3v^2 a= \frac{(100 kg)(10 \frac{m}{s^{2}}) - 0.3v^2}{m} \frac{dv}{dt}= 10 \frac{m}{s^{2}} - 0.003v^2 \frac{dv}{10 \frac{m}{s^{2}} - 0.003v^2}= dt...

I integrated to find the answer, which lead to about 17.55 seconds.

Given a situation where an object is dropped at a certain height, how should the net force equation be written considering the fact that only gravity and drag acts on it? I'm stuck on whether the equation is written like this: \Sigma F = F_{g} - F_{d} Or like this because g is negative?: \Sigma...

I redid the problem on paper this time and I got a slightly different answer: h=\frac{2mg}{pCA} \times ln|\frac{e^{2 \sqrt{\frac{pCAg}{2m}}t}+1}{2e^{\sqrt{\frac{pCAg}{ 2m}}t}}| Compared to my original one from the above post: h=\frac{2m}{pCA} \times ln|\frac{e^{2...

Because the object is at rest at time 0, then the object has moved a distance of 0: \frac{\beta}{\sqrt{g}}x=\frac{2 \alpha }{\beta} \times ln|\frac{e^{\frac{\beta}{ \alpha }t}+1}{2e^{\frac{\beta}{2 \alpha }t}}| + C \frac{\beta}{\sqrt{g}}0=\frac{2 \alpha }{\beta} \times ln|\frac{e^{\frac{\beta}{...

Continued from above: \sqrt{g}+v\beta=e^{\frac{\beta}{\alpha}t} \times (\sqrt{g}-v\beta) \sqrt{g}+v\beta=e^{\frac{\beta}{\alpha}t}\sqrt{g}-e^{\frac{\beta}{\alpha}t}v\beta e^{\frac{\beta}{\alpha}t}v\beta+v\beta=e^{\frac{ \beta }{\alpha}t} \sqrt{g}-\sqrt{g} v\beta \times...

So it should be solved like this?: \frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} \times ln\left(\frac{|\sqrt{g}+v\sqrt{\frac{1}{2m}pCA}|}{ |\sqrt{g}-v\sqrt{\frac{1}{2m}pCA}|}\right)=t+D Because the object is at rest at time 0, then D is...

Can someone confirm if I manipulated the hyperbolic function correctly and if I integrated correctly?

Well I managed to manipulate and simplify the equation into this, but I haven't learned how to use hyperbolic functions, so what do I do? a = \frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} b = \sqrt{\frac{1}{2m}pCA} \frac{a}{\sqrt{g}}\frac{ds}{dt} = \frac{e^{bt} - 1}{e^{bt} + 1}...

So now that I've solved the velocity function, is there a way to integrate to find the position function of time?

Homework Statement Given a situation where a ball of mass m is dropped off at a height of h, find the equation that would give the velocity of the ball with respect to time given that gravity and drag are intrinsic factors. Homework Equations F_{g} = m g F_{D} = \frac{1}{2}pCAv^2...

Shouldn't the force be 50 N? Since if the work is 10 J, and the distance traveled is 0.2 meters, then 10 J / 0.2 meters = 50 N. And if 50 N is the force then the power is equal to force times velocity, 50 N * 8.2 m/s, which is equal to 410 J/s?