Recent content by Kampret

yes it just when i try both method it doesn't display same result so it leave me restless since i think both are valid equation but I'm not sure about the second since so far in problem lIke this is always give either mass of area just like what haruspex stated above . frankly it first time i...

i don't say any type of triangle. if you open the link you will see it only ordinary triangle

Homework Statement if the black dot is assumed by (0,0).find the center of mass coordinate of this triangle [/B] i'm sorry but since the pic won't show ill attach the link here https://ibb.co/4Ptw5T7 <Moderator's note: picture added> Homework Equations centroid is 2/3 of median [/B] using...

thanks, that was truly beautiful! . i guess keeping the variable as short as possible until the last equation really help to prevent confusion...

Well, I've been try my best to solve that problem and i do not thought that i made mistake so I've been restless these few days wondering how solve this, and if the problem is really the one which wrong i am really glad iam worried over nothing , but if that value $$ x = \sqrt{vv0}t$$ value...

i like to apologized, i do not intend to do such behaviour , i am also do not know that something wrong in this question and i can't prove the answer by myself so i tried brought my question here, what i typed here is exactly like written in the book. once again please forgive me

thank you it's is really reassuring hearing it from you, in addition i want to ask about my final result (post number #12) and the process of it is written on post number #9 what do you think about it ? i've been restless these past few days thought about this problem since my math is not very good.

i'm sorry if my question seems little foolish, is that mean $$-{4 \over k(kt + c_1)}$$ is equal to $$\sqrt{vv_0}$$ and $$2 \sqrt{v_0} \over kt + c_1$$ if yes would you mind to tell me how $$-{4 \over k(kt + c_1)}$$ turn into $$\sqrt{vv_0}$$ and $$2 \sqrt{v_0} \over kt + c_1$$ I'm trying to...

after take a look at gneill's hint i realized that i made a mistake in the end before, i write like this : $$ x = \frac {-2} {k} v^{0,5} + \frac 2 k v^{0,5}$$ but it should be $$ x = \frac {-2} {k} v^{0,5} \frac {-2} k v^{0,5}$$ so it will turn into $$ \frac {xk} {-2} = \sqrt {vv0}$$ it seems...

$$\frac {dv} {dt} = a$$ $$\frac {dv} {dt} = -kv^{1,5}$$ $$\frac {dv} {v^{1,5}} = -k {dt}$$ $$\ {v^{-1,5}}{dv} = -k {dt}$$ $$\int {v^{-1,5}}{dv} = - \int k{dt}$$ $$\frac {v^{-0,5}} {-0,5} = -{kt} + C$$ $$\ {-2} {v^{-0,5}} = -{kt} + C$$ Now for the initial condition when t=0 v=v0 $$\ {-2}...

thanks , i will check it later

I'm sorry for late reply and thank you for your help but any some part of your writing is misplace when it should be nominator it become denominator in yours , so I'm on my way construct it and once i finish i will post it here

i'm truly sorry but i never use latex before, so I'm not sure from where i should start, and since it quite complicated i will try learn it from the basic on very near future

Homework Statement acceleration of moving particle is described by a=-kv^1,5 where k is a constant. if the condition when t=0 is v=v0 and x=0 prove that xt = √(vv0).t Homework Equations dv/dt=a, dx/dt=v The Attempt at a Solution dv/dt=a dv/v^1,5=-k dt v^-1,5 dv = -k dt ← integrating both...

ok, so what should i do to integrating v so far i only faced problem like this √(h²-x²)dx where h is constant so i can turn x into sin and dx into cos. but this just worked because there is both x and dx in that equation but here i encounter like what ray's wrote v=√(c²-1/9x²) so for the...