physic question? piease write the way of solution.?
442329
there would be many answers. a lot of combination of speeds and angles are possible to put the ball in the basket.
855447
if it is equations of motion then \nv=u+ft\ns=ut+1/2ft^2\nv^2=u^2+2fs\nhope this is what you wanted
935341
1 sum the forces and set equal to zero. 2 forces, gravity and electric.\n\n2. conservation of energy. initially all kinetic, finally all potential. KE1=U2. U2 is electric potential. solve.\n\ngood luck
179131
Try this:\nhttp://kr.cs.ait.ac.th/~radok/physics/f3.htm
1116027
Find the potential energy PE=m*g*h\nEquate it to the energy equation\nMass*Height*accln due to gravity=Rise in temp.*Mass*Sp.Heat Cap.\nCancelling mass from both sides and rearranging the equation.\nRise in temp=Height*accln due to gravity/Sp heat capacity\n                    =(200*9.81/5.04*10^2) C
397497
I would use the ideal gas law...more specifically...the combined ideal gas law...\n\nP1*V1/T1 = P2*V2/T2
388135
since it is difficult to use the equation writer on Yaho answers, I am giving you a link, that has the answer.\nhttp://homepage.mac.com/nshoffner/nsh/CalcBookAll/Chapter%206/Ch6DefInt.htm\nScroll down a little.
805064
lol \nYou have homework that you don't know the answer to.  \n\nWhy would someone ask these here out of interest? Like you have these specific parameters and conditions that you are just raring to want to know just out of the blue.\n\n Quit wasting time writing and typing questions out waiting here for answers when you could be the next engineering graduate who actuallly UNDERSTANDS this stuff by doing it yourself.
52130
E=mc∆T is the equation you want to use for this situation, where E is energy, m is mass, c is the specific heat, and ∆T is the change in temperature.  Now, the heat lost by the alloy is equal to the heat gained by the water, right?  So look what we can do:\nmc∆T=mc∆T\nwhere the left side of the equation is talking about the metal, and the right side of the equation is talking about the water.  Let's start pluggin in values.  We'll leave the left c value as a variable because that's what we're solving for.  \n36.0 g *c*(178 oC -30.6 oC)=185.0 g*(4.184 J/g*K)*(30.6 oC-25.0 oC)\nThe trickiest thing about that step is the temperatures: the metal started at 178 and goes down to 30.6; the water starts at 25 and goes up to 30.6.  \nNow evaluate that equation and solve for c. \nc=.817 J/g*K
991058
mass =concentration\n-------\nvolume
974937
please complete the equation
969486
I just helped someone else with their physics question, so I'm alittle tired of it.  But, I suggest you make an equation sheet...physics is all about equations.  So, if you have a piece of paper with your equations, all you have to do is look at the sheet and plug things in...
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
1132678
Directions how to do this can be found in the text book.\n\nDo your own homework!
942436
F(t-t0) = m(v - v0)\nF = 0.25W = 0.25mg\n0.25*9.80662*t = 8 m/s - 5 m/s\nt = 3/(0.25*9.80662) s\nt = 1.2237 s
48084
F = ma    \nNewton's second law.
715037
Ok-i'm not quite sure if i did this right--we covered this last year--so you might want to double check with your textbook or teacher, or whatever resource you have...\n\nwhat i would do is find the final temperature:\nFinal Temperature=(V2 * T1) / V1\n\nand then use the delta T, delta v, and P to find delta E.\n\nbut i'm really not that sure.\n\nsorry that i wasnt a better help!