what formula do I use to figure this physics question out?
136906
pH = -log[H+]\n\nand\n\n[H+][OH-] = 1*10^-14\n\nwhere [ ] means concentration of.\n\n\nSo, to work out [H+], you divide 1*10^-14 by 3*10^-10 to get  3.33333*10^-5\n\nTake the -log of this, and you get 4.477.\n\nThe pH is 4.477.\n\nAll logs are in base 10.
449781
There are jthree  formulae cncerning velocity.These are v=u+at , V^2=u^2+2as ,and s=ut +(1/2) at^2 ,where u= initial velocity,v=final velocity ,a=accelaration.,s= distance travelled and t= time.Now how to decide which formula is to be used?in the problem,just see what they want you to find out and note down what is are already given in the problem, if they give you u,a and t and ask you to find out v,apply formula no.1.But if they gdonot give you time but give the distance or displacement and asks you to find v,donot hesitate to apply formula no,2 and so on.if you have any difficulty to understand.please do not hesitate to write to me through yahoo answersand i shall be glad to assist you
105407
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
490170
A) I think I got it...Use energy, except for g, use   Fg cos(19 degrees), found from a free-body diagram.\n\nso 1/2 mv^2 = mgh\n\n1/2 (4.7)^2 =  9.8(cos (19))  h    solve for h\n\nB)  now that you have h...maybe you can use   \n\nd y = V(sub 0 sub y) t + 1/2 (a x cos (19)) t^2\n\nyea, tell me how it comes out.
394261
Hi, \n\nIn general, the equation describing the position of an object under the influence of a constant force is as follows:\n\nh(t) = at^2 + vot + xo \n\nwhere : \nh(t) = height at any time t\na = acceleration\nvo = initial velocity\nxo = initial height.\n\nSo the answer would be \nF= m*a (mass times accelleration) = (the mass kg)*(1.96m/s^2) \n\nI'm guessing that you want to find out the equation for acceleration, so what you should do is use the values for t, and   h(t) to solve for the constants (xo, vo)\n\nexample: \nat time t=0, height is 0.50m\n\nso h(0) = a(0)^2 + vo (0) + xo = xo\n\nbut that also equals 0.50m;\n\nso:\nxo = 0.50m.\n\nand the equation becomes \nh(t) = at^2 + vot + 0.50m. \n\nThen use the different values of t, and h(t) to do the same thing with the other constant, and you'll have it. \n\nHope that helps, \nMatt
353952
This problem is a conservation of momentum problem:\n\n(a) In an elastic collision, the formulas for calculating the final velocities are derived from both the conservation of momentum formula and the conservation of kinetic energy formula. Check out my source for more information, but for now just trust me that they are valid:\n\nv1 = (v1(m1 - m2) + 2m2v2)/(m1 + m2)\nv1 = (1*(4 - 7.2) + 0)/(4 + 7.2) = -0.3 m/s\n\nv2 = (v2(m2 - m1) + 2m1v1)/(m1 + m2)\nv2 = (0 + 2*4*1)/(4 + 7.2) = 0.7 m/s\n\n(b) In an inelastic collision, just use the conservation of momentum formula:\n\nm1v1' + m2v2' = (m1+ m2)v\n\n4*1 = (4+7.2)*v\nv = 4/(4+7.2) = 0.36 m/s\n\nI hope this helps!
542542
Given: m=1.6*10^7kg\n           F=8.5*10^5N\n\nI only guessed that the above is what you meant.\n\nSolve for a, using the formula:\n\nF=Ma\n\n8.5*10^5=1.6*10^7a\na=8.5*10^5/1.6*10^7\n  =5.31*10^-2m/s^2\n\nv=70km/h =70*1000/3600m/s\n  =19.44m/s\nu=0\n\nTo solve for time, t, use the formula:\n\nv=u+at\n\n19.44=0+5.31*10^-2t\nt=19.44/5.31*10^-2\n =3.66*10^2\n =366s
422571
Using one of the equations of motion\n\n(v*v) - (u*u) = 2*a*s\n\nwhere \nv is final velocity\nu is initial velocity\na is accleration\ns is displacement\n\n\nOR \nuse \n\nv * (dv / ds) = a
470009
Use the certical component to find the time in the air and since the horizontal velocity does not change; Dx = vx * time\n\nDx = v^2 / g\n\nheight = v^2 / (2g)\n\nHope this helps without actually solving it for you.
511144
You're on the right track\nOne small clarification:\n\n\nm1gh = 1/2m1v1^2 + 1/2m2v2^2\n\nm1gh=5*9.8*5\n=245\n\nYou can also calculate the velocity of m1 at the moment of collision:\nm1gh=1/2m1v1^2\n=sqrt(9.8*5*2)\n=9.9\nso\n5*9.9=10*v2f+5*v1f\n9.9=2*v2f+v1f\nv1f=9.9-2*v2f\nSubstitute into the quadratic:\n245*2=5*(9.9-2v2f)^2+10*v2f^2\n245*2/5=\n9.9^2-4*9.9V2f+4V2f^2+2v2f^2\n6v2f^2-39.6V2f+.01=0\nSolve\nfor v2f\n0, 6.6\nusing 6.6\nv1f=9.9-2*6.6\nv1f=-3.3\nagain, \nusing mgh=1/2mv^2\nh=.5*4.4*3.3/9.8\nh=.74m\n\n\nj
378002
first get the time from the horizontal motion  t = .51 / 1.2 s (uniform motion, no acceleration here)\n\nthen use h = 1/2gt^2 for the vertical motion (with initial velocity zero)
645938
in order to solve this problem, you need to work part (b) first before (a)\n\nb): you know gravity (g=32.2 ft/sec) acts opposite the rocket, so calculate, using:\n\nv_final = v_initial - 2*g*y_1\n\nyou know that the final velocity is zero (max height) and you are given the inital velocity. the negative sign is to account for the opposite direction of gravity. move the equation around to solve for y_1 (height from tower to max. height). then add 100 feet (to account for the tower) and you have your answer.\n\na): since you know your maximum height (from part b), substitute that answer for h. [*NOTE: remember to subtract 100 feet from your answer in part b, because t=0 at the top of the tower] You now have a quadratic equation and solve for t using the quadratic formula (you should know what that is). your answer will be the answer!\n\nThis prob is easy only because it worries upto the max height. if we had to worry about the falling down part, well thats another story.\n\n** i already worked out this prob, so if you want to ask more questions/verify your answer, just post an additional response and ill modify my question to answer it. GL!
590413
F = Gm1m2/ r^2\nF = ma  \na = v/t\nv = d/t\na =d/t^2 \nF = m d/t^2  = Gm1m2/ d^2\nm  = (m1 + m2)\nF = (m1 + m2) d/t^2 = Gm1m2 /d^2\nd/m2t^2 + d/m1t^2 = G/d^2\nd/m2 + d/m1 = Gt^2/d^2\nSQRT{ (1/G) d^2 (d/m2 +d/m1)} = t
521012
calculate the angular acceleration of the seed:\nw_i = 0\nw_f = 3.49 rad/sec\n\nalpha = (w_f-w_i)/t\nalpha = (3.49-0)/0.25\nalpha = 13.96 rad/sec^2\n\nconvert this to linear acceleration at the seed:\na = alpha*r\na = 13.96*0.060\na = 0.8376 m/sec^2\n\nThe maximum centripetal acceleration occurs when the record is moving at max speed so find the max centripetal acceleration:\na_c = (w_f^2)*r\na_c = (3.49^2)*0.060\na_c = 0.731\n\nthe 2 accelerations are perpendicular to each other so find the magnitude using Pythagorean:\nsqrt(0.731^2 + 0.8376^2) = 1.112 m/sec^2\n\nthen calculate the net forces:\nu*N = m*A\nu*(m*9.8) = m*1.112\nu = 1.112/9.8\nu = 0.113
473517
If you're using m to indicate the mass, the differential equation should be:\ndm/dt = km\nIf you're using y, then:\ndy/dt = ky\n\nThe error in your answer is that you are alternately using two variables, y and m, to indicate the mass. Pick one and stick with it.
510144
use the formula\nA=2pir(h+r)\n=2pi*7*21\n=294pi
329247
A lot of physics (especially beginning physics) is just finding the right equation and plugging in what numbers you are given.\nEquations usefull for your problem:\nKE = 1/2*m*v^2 (KE = kinetic energy, m = mass, v = velocity)\nAND\nWork = Force * distance     This is a very important equation\nit's easy to memorize so do it!