µs = Ms ~  Mk (?)     and     Mk = f (v,m) (?) ... How do you interpret this equation?
143456
I think "µs" and "Ms" are both the coefficient of static friction; they're just in different notations. "~" means "is similar to". The first equation is just saying that the coefficient of static friction and coefficient of kinetic friction are similar quantities: static is for non-moving contact surfaces, and kinetic is for moving contact surfaces.\n\n"Mk = f (v,m)" just means that the coefficient of kinetic friction is a function of both mass and velocity. This means that the coefficient is dependent of both mass and velocity (if either of them change, the coefficient will change also). It's important to note that mass and velocity are not dependent on each other.
798483
Here it is:\n\n(98,42)Mo + (1,0)n ---> (99,42)Mo
987931
you are making a mistake with the units\n\ndid you mean kg/s^2? if it is so the answer is 10300 kg/s^2\nif did you mean kgf/m it is 1050 kgf/m \nif not your units are wrong. \nremember that 1 N (newton) is 1 kgxm/s^2\nand 1 kgf = 9.8065 kgxm/s
250190
This is really simple stuff. Newtons is force, seconds is time, meters is length, and meters per second is length per time, or velocity.
697157
False.  m/s (meters per second?) is not a measure of rotary speed.  Until you get that straightened out, no answer can be calculated for a) or b)\n\nBefore posting a ?, do a search here.  This same ? came up yesterday.........!
728450
G=(T+Mf)/M \ncross-multiply\nGM = T + Mf\nGM - Mf = T\nfactor out the M\nM(G - f) = T\nM = T(G-f)\n\nANSWER\nM = T(G-f)
888459
v(t)= -gt - ve(ln((m - rt)/m))\n\nx(t) = ∫v(t) dt\n\n= ∫[-gt - ve(ln((m - rt)/m))] dt (Note ∫lnu du = ulnu - u + k)\n\nLet u = (m - rt)/m Thus du = -r/m dt ie dt = -m/r du\n\nSo x(t) = -½gt² - ve*-m/r (u(lnu - 1) + c\n= mve/r [ (m - rt)/m * (ln((m - rt)/m)  - 1] - ½gt²\n= ve(m - rt)/r * (ln((m - rt)/m) - 1) - ½gt²\n\nIf g = 9.8 m/s^2, m = 30000 kg, r = 155 kg/s, and ve = 3000 m/s and t = 1 minute = 60 s\n\nx(60) = 3000 * (30000 - 155 * 60)/155 * (ln (30000 - 155 * 60)/155 - 1) - ½ * 9.8 * 60²\n\n≈ 138390 m\n\n= 138.39km
147821
- -----------O\n------------ . .-----------  _\n  K+ ( O: Mn  :O )\n-------------. .\n------------ O\n\nomit the long dotted lines
34572
alright, well just convert the mN into N and divide by gravity... \nso we multiply be the conversion factor getting .00000671N. now F=ma=mg so m=F/g(mass equals Force divided by gravity)  so .00000671/9.8= 6.85 x 10^-7 kg
954813
Yeah, it's E=mc^2 and it relates energy and mass.  E is energy, m is mass, c is the speed of light (celeritas).  The equation could come useful in physics, like atomic physics.\n\nWhat did you wanted to know?
502184
5H2O2(aq)+2KMnO4(aq)+\n3H2SO4(aq)-->\n2MnSO4(aq)+K2SO4(aq)+\n8H2O(l)+5O2(g)
48084
F = ma    \nNewton's second law.
698608
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
733015
a. Applying conservation of momentum\nm1 x V_i + m2 x 0 = (m1+m2) x V_f\nV_f = (m1 x V_i)/(m1+m2)\n\nb. If m1 >> m2 then m1 + m2 -> m1\nTherefore V_f = V_i\n\nc. If m1 << m2 then m1+m2 --> m2\nTherfore V_f = (m1/m2) x V_i\n\nd.m1 x V_i - m2 x V_i = (m1+m2) x V_f\n=> V_f = ((m1-m2)x(V_i))/(m1+m2)\n\ne. m2 = m1, m2 =(1/3)(m1)
700987
The energy taken to slow the object from v1 to v2 is\n\nE=FS=(1/2)m(sqr(v2)-sqr(v1))\n\nand given v2=(1/2)(v1)\n\nWe get  v1=sqrt((8/3)(FS/m)) = 12m/s\n\nAlso from v2=v1+at     and          F=ma\n\nt=v1/(2a)=mv1/(2F) = 1 second
1044611
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
958486
Rs=Im*Rm/Is\n\nRs*Is = Im*Rm\nRm= (Rs*Is)/Im\n\nc=(8.85*10^-12)(K)(A/d)\nc*d = (8.85*10^-12)(K)(A)\nd= [8.85*10^-12)(K)(A)]/c\n\ny=mx+b\nb=y/(mx)\n\nx= (y-b)/m\n\nLm=(Lt-L1-L2)/2\n\nE=Q/C\n\n1/Rt - 1/R2 = 1/R1