## anonymous 4 years ago Dimensional analysis. I'm extremely confused :/. 1. Check which of the following equations are dimensionally correct: x=(X_not)+(v_not)(t)+(1/2)(a)(t^2). Not sure if it looks right when I type it out.

1. anonymous

This might make it easier to understand. I am just trying to figure out how to do this. I'm really confused and my first lab exercise is tomorrow :/

2. anonymous

x is in meter, $$\mathsf{x_{0}}$$ is in meter as well, $$\mathsf{v_{0}}$$ is in $$\mathsf{ms^{-1}}$$, time in s and acceleration (a) in $$\mathsf{ms^{-2}}$$. Note. I am assuming all the quantities are in SI units. $\mathsf{ m = m + ms^{-1} * s + \frac{1}{2}*ms^{-2}*s^{2} }$ I haven't read your question, but I think this is what you are supposed to prove.

3. anonymous

The equation is Dimensionaly Correct!

4. anonymous

thank you Ishaan. Just to clarify, how do I know which symbols are dimensional units? is x always meters and v always velocity? Also, I like your photo :). Anonymous is awesome.

5. anonymous

ok I think I understand now :D. It just looked really scary at first because of all the symbols, but it's really just basic algebra right?

6. anonymous

I know, thanks! @ anonymous Yeah, just basic algebra.

7. anonymous

Thanks man. I feel so slow sometimes xD. I'm a technophile, and yet when it comes to basic physics and math, if something looks uncomfortable to me, I freak out... it's like a mathematical learned helplessness haha.

8. anonymous

I really wish I had a better sense of spatial reasoning though... I can't even do multiplication or addition in my head if it's above a factor of 12 xD haha

9. anonymous

ok I think I might be stuck on part D of #1. Do I need to take a natural log or something?

10. anonymous

Even I can't; you need not worry about calculations just focus on the concepts that's it, for all the calculations we have calculators.  $$\mathsf{1.D}$$ $$\mathsf{\text{KE} = \frac{1}{2}.m.\frac{x^2}{t}}$$ $\mathsf{\text{KE} = \frac{1}{2}.m.v^2}$ Now just put in the units, you don't need logarithm or anything else. $\mathsf{kg*m^2 s^{-4} = KE}$ $\mathsf{\frac{1}{2}.m.x^2.\frac{1}{t} = kg*m^2*\frac{1}{s}}$ Hence, this equation isn't dimensionaly correct. Note. kg for mass (m), $$\mathsf{ms^{-1}}$$ for velocity(v) and s for time (t).

11. anonymous

ah ok. thank you so much. I think I know how to do the rest of these questions now. Everyone I've met on this site is so nice. You guys are really altruistic, and you provide your knowledge to others for free. Much respect :)

12. anonymous

Yeah, OpenStudy platform is really nice. You should checkout the mathematics section it's the most popular section on OpenStudy.

13. anonymous

yep, they've helped me a lot too. Anyway, I won't bother you too much now :p. Thanks for everything. If I get stuck in the future, you'll most likely be the first person I ask for help (if that's ok with you) :p haha.