mcastillo356's latest activity

Welcome to PF! You've come to the right place. :smile:

At first glance, ##m\vec{g}=\vec{F_n}## But I don't know how to solve it if I must consider geometry. ##V=\cfrac{1}{3}\pi\,r^2\,h##...

I think you may have a misunderstanding of what "peer review" means, though it may partly be moot.... "Peer review" is what happens...

Right. In your step 3, you ended up with 99, which is the same as 099. Reversing the digits of the latter 3-digit number results in 990...

How does my arithmetic end up with ##1089##?. I don't understand. Maybe missing something of the statement?. Attempt The statement...

If I have understood, your example ends up with ##999## Counterexample 1- ##495## 2- ##594## 3- ##594-495=99## 4- ##99## 5-...

Well you rumbled into PF like a bull in a china shop, what do you expect?! :smile:

Thanks to everyone who voted for me. I must confess I feel more like a leg-end than a legend, but there you go.

Last step is this way $$mc^2t^{-1/2}\bigg |_{1}^{1-\frac{u_f^2}{c^2}}=mc^2\left (\frac{1}{\sqrt{1-u_f^2/c^2}}-1\right )$$

My special thanks to Fresh42, Mark44, Bvu, Perok, SammyS, Haruspex, Orodruin, Chestermiller, WWGD, Martinbn, and the other person from...

$$\displaystyle\int_0^{u_f}m\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du=m\displaystyle\int_0^{u_f}\Bigg (1-\cfrac{u^2}{c^2}\Bigg...

An integral is an antiderivative. The only thing you need to check is that: $$mc^2\frac d {du} \bigg...

You can also derive ##d \gamma = \gamma^3 (\vec \beta \cdot d \vec \beta)## fairly easily and use that to do a change of variables...

Good afternoon, This is not Paul, but rather his son Robert. I am very sad to write here that my dad passed away on Saturday...

I think this is the integral you are attempting to compute. If so, I'd do it by inspection. For any integrand with a ##1/f^n(x)## it's...