I cant derive 3rd equation of motion i.e. V raise to power 2=\nU Raise to power 2 +2AS
205010
Which graph du you use? velocity verses time or velocity versus distance?
342065
Have you been given the standard formulas?\nx = a cos (nt + c),  frequency = n, period = 2*pi/n, phase = c (Not sure of that -- it may be -c.  Look it up.) This gives you the answers to (d), (e) and (f) immediately.  [No, I'm not doing it for you: You can substitute values I'm sure.]\n\nx is the displacement, so just sub t = 2 in this equation.  If you need touse your calculator (you shouldn't!), make sure it's set in radians.\n\nVelocity is dx/dt.  You can do that I hope, and get\n\nv = - 18*pi*sin(3*pi*t + pi/3)  Notice I omit units.  I prefer to use equations relating to numbers only (though I know units are sometimes useful especially in a Physics context) and then state the correct units in presenting the answer -- which you are going to do!\nSub t = 2 to get v in m/s\n\nAcceleration is dv/dt, so differentiate and then sub t = 2\n\nAll the best
63853
V = I * R = I1 * R1 = I2 * R2.\n\nI2 * R2 = I1 * R1\n\nI2 = (I1 * R1)/ R2\n\nR2 = 16 * R1\n\nI2 = (I1 * R1)/ (16 * R1)\n\nI2 = I1 / 16
488300
its original velocity is in x direction   use v=vo+at for each coordinate\nthen  vx= 152+8.92x3.10\n         vy=   0  +1.21x3.10\n v=sqrt(vx^2+vy^2)
387575
p= 2L + 2 W \n2w=p-2l\nw=(p-2l)/2
407585
Full Answer Here! :))\n\na) s=u(v(x)) = -2(v(x))^2+2(v(x))+1 = -2(3x-1)^2+2(3x-1)+1 = -2(9x^2-6x+1)+6x-2+1 = -18x^2+6x-2+6x-2+1 = -18x^2+12x-3.\n\nb) t=v(u(t)) = 3(u(t))-1 = 3(-2t^2+2t+1)-1 = -6t^2+6t+3-1 = -6t^2+6t+2.\n\nQuality ia 100%!
359931
Its substitution.\n\nYou are given p=i*v\n\nand you need to express it in terms of one variable using ohms law v=i*r\n\nYou want p, so you need to trade out either the i or the v from p=i*v\n\nif you trade out the v using ohms law as (i*r) then your power equation becomes p=i*(i*r) = i^2 *r\n\nif you trade out the i using ohms law, first you solve for i getting i=v/r then substitute into the p-equation.  \nThe substitution yields p=(v/r)*v  = v^2 /r\n\nThis is straightforward algebra.
594639
start with   d = m / v        multiply both side by v; \nyou get     v * d = m        divide both sides by d; \nyou get     v = m / d
366293
you gotta integrate it, i forget what way it works but my guess is work is the integral of the force so do integral from 2 to 3 of 4x+3\nim sure you have to integrate just not sure which way check it up
16477
The answer above is WRONG. You can tell it is wrong because if you set m3 = 0, the mass has an infinite acceleration, which is unphysical (impossible)\n\nYou have to apply Newton's second law to all 3 masses in the system.\n\nm1*a = m1*g - T1\nm2*a = T2 - m2*g\nm3*a = T1 - T2 - m3*g*mu (mu = coeff of friction)\n\nAdd all 3 equations together and rearrange to get \na = g*(m1 - m2 - m3*mu)/(m1 + m2 + m3)\n\nWork the rest out by substution into the original equation.
476030
we have V*T = const\nV1*T1 = V2*T2\n=> V2 = V1*T1/ T2 = 3.8 * (-45 + 273) / (45 + 273) = 2.72L
344253
Hello, I would be glad to help you with this:\n\n a=(v-c)/t\nat=v-c\n-c=at-v\nc=v-at \n\n I hope this helps and good luck with the rest of your course work ,Please Please remember me for best answer
497614
a= r * w^2 = .12 * 2^2 = .12 *4 = .48 m/s^2
202051
P1 x V1 / T1 = P2 x V2 / T2\n\nYou have all you need.  Calculate it.
469860
a) As the movement is circular, the total of the forces is in this plan and to the centre and the value is mw²R.\n\nSo P + S = mw²R where P is the weight and S the tension of the string and R the radius and a the angle between the string and the vertical\nIn projection, we have :\nmw²R = S sin a\nmg = S cos a\n\nDividing the 2 equations :\ntg a = w²R/g\nand R = L sin a\n\nso w² = g/(Lcos a)\nso cos a = g/Lw² = g T²/(L 4 pi²) \n\nso w²= g/(L cosa)  \n\nso v = r sqrt(g/(Lcos a)) = 1.4 sin 30  sqrt( 9.8/1.4 cos30) = 1.99 m/s\n\nb) 4 = 1.4 sin a sqrt (9.8 /1.4 cos a)\n\nc) As mw²R = S sin a\n    and mg = S cos a\n\nw = sqrt (S²- m²g²) /mR
421860
here, you basically use the kinematics equations:\nx=x0+v0cos(theta)t+.5axt^2\ny=y0+v0sin(theta)t+.5ayt^2\nhere, you can set x0 to be 0 since you're only interested in the change in x.  also, there's no acceleration in the x direction, so ax=0.  you start out at y=0, and after the jump, you end up at y=0 again:\nx=v0cos(theta)t\n0=v0sin(theta)t-.5gt^2\nv0sin(theta)t=.5gt^2\nv0sin(theta)=.5gt\nt=2v0sin(theta)/g\nx=v0cos(theta)*2v0sin(theta)/g\nnow, you must use the trig identity 2cos(theta) *sin(theta) =sin(2*theta):\nx=v0^2*sin(2*theta)/g\nthis formula is known as the range equation.\ng=v0^2*sin(2*theta)/x\nhere, you have 2 unknowns (g and theta), but you do know that for g to be at a maximum, sin(2*theta) must also be at a maximum.  since sin(2*theta) has a maximum value of 1, which occurs at theta=45 degrees, you have:\ng=v0^2/x\n  =(5.90 m/s)^2/7.52 m\n  =4.63 m/s/s
590930
Momentum - P = mv (P=momentum, m=mass, v = velocity)\nThree Laws of Newton - F = ma (F=Force; a=acceleration)\nEquations of motion - 1) v = u + at\n                                    2) S = (1/2)ut + at^2\n                                    3) v^2 = u^2 + 2aS\n(v=final velocity; u = initial velocity; S=distance)\n\nIf this can help you please mail me back !\nBEST OF LUCK !!!!