Derive Us=tan (theta) (apply 2nd law of motion for constant velocity)?
111071
I think you are referring to the coefficient of static friction u(pronounced as myoo)s. This involves some derivation of equations but if you'll review your static friction on inclined planes, you can easily start from there.
382082
use d=V1*t + [(a*t*t)/2]\nd=distance=490m\nV1=initial velocity = 0m/s\nt=time = ?\na=acceleration due to gravity = 9.81m/s*s\n\nusing equation above, t will come up to be:\nt=(square root)(2*d/a)\nt=(square root)(2*490m / 9.81m/s*s)\nt=9.99 sec or 10 sec.
1001657
V= V(i) + AT Initial velocity plus acceleratin x Time\n\n\n   = V(i) + (F/m)T     Acceleration = Force/ Mass\n\nNote  use same system of units
46701
v=u+at\nv^2-u^2=2as\ns=ut+1/2at^2\ns=disance traveled\nu=initial vel.\nv=final vel.\na=accelaration\nt=time
610143
I think ur right cos^2(theta) should be cos(theta). then the answer would be perfect.
666681
1/2 * mass * velocity^2.
861830
The formula for distance traveled under constant acceleration a is s=.5*a*t^2.  The general formula is a(t) = d^2 x(t) /dt^2, acceleration is the second derivative of position (distance) with respect to time.
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
817617
tan(A-B)+tan(B-C)+tan(C-A)\n= (tanA - tanB)/(1 + tanAtanB) + (tanB - tanC)/(1 + tanBtanC) + (tanC - tanA)/(1 + tanAtanC)\n= [(tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC)]/[(1 + tanAtanB)(1 + tanBtanC)(1 + tanAtanC)]\n= 0\nie (tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC) = 0\nOn expanding\n(tanA - tanB)(1 + tanAtanC + tanBtanC + tanAtanBtan²C) \n+ (tanB - tanC)(1 + tanAtanC + tanAtanB + tan²AtanBtanC)\n+ (tanC - tanA)(1 + tanAtanB + tanBtanC + tanAtan²BtanC)\n\n= tanA - tanB + tan²AtanC - tanAtanBtanC + tanAtanBtanC - tan²BtanC + tan²AtanBtan²C - tanAtan²Btan²C \n+ tanB - tanC + tanAtanBtanC - tanAtan²C + tanAtan²B - tanAtanBtanC + tan²Atan²BtanC - tan²AtanBtan²C\n+ tanC - tanA + tanAtanBtanC - tan²AtanB + tanBtan²C - tanAtanBtanC  + tanAtan²Btan²C - tan²Atan²BtanC\n= tan²AtanC - tan²BtanC - tanAtan²C + tanAtan²B - tan²AtanB + tanBtan²C\n=0 \nDivide all by tanAtanBtanC\nSo tanA/tanB - tanB/tanA - tanC/tanB + tanB/tanC - tanA/tanC - tanC/tanA = 0\n\nie tanA/tanB + tanB/tanC + tanC/tanA = tanB/tanA + tanC/tanB = tanA/tanC\n\nWhence tanA/tanB + tanB/tanC + tanC/tanA = cotA/cotB + cotB/cotC + cotC/cotA\n\nSince a triangle then tan(A + B) = tan(π - C) = - tanC\nSo (tanA + tanB)/(1 - tanAtanB) = -tanC\nie tan A + tan B = tanAtanBtanC - tanC\nso tanA + tanB + tanC = tanAtanBtanC
850884
using chain rule and product rule\n\ny'= 3x^(2)*tan(4x)\n         +4x^(3)*sec^(2)(4x)\n  = x^2(3tan(4x)+4x*sec^(2)(4x) \n\ni hope that this helps
918000
Ok, iam assuming you know the equations of motion: \nv= u + at\ns=ut + 0.5at^2\nv^2=u^2 + 2as\ns=((u+v)/2)t\n\nWe need to break this motion up into seperate sections.\n\nNote however that all the units in this question are in SI units, so they are consistent with each other. You don't have kilometres which you would need to convert into metres for example, before using these equations. In that case I won't put in the units by the numbers in my calculations everytime, just to make it look clearer. But they are there and they are consistent.\n\n\nFirst section is from rest (u=0) accelerating at 2.6m/s^2 for 18s. We will find the displacement travelled here.\n\nWe know u(initial velocity)=0, we know a(acceleration)= 2.6 and t=18 and we want s(displacement)\n\nthe equation connecting these four is s=ut + 0.5at^2 \n(we don't need v and this equation does not have v)\n\nso s = (0 x 18)  +  (0.5 x 2.6 x 18^2)\n       = 421.2m << note this displacement, we're going to use it at the end\n\n\nThe next section, we have the automobile continuing its motion at a constant velocity that it has gained after accelerating at 2.6m/s^2 for 18s.\n\nHere, we need to find what this constant velocity is, which is easy. \n\nSince the body is accelerating at 2.6 metres per second every second it is gaining a velocity of 2.6 metres per second, every second.\n\nSo the velocity it has gained after it has accelerated for 18s is just \n2.6m/s^2 multiplied by 18s \n= 46.8m/s^1\n\nIt's important you notice that I am using 18s because this is how much time it has accelerated for to reach that constant velocity; it stopped accelerating after 18s. 15s is the time it travels at that speed it has gained after accelerating for 18s.\n\nNow this velocity is constant so every second it travels a displacement of 46.8m, the displacement it travels here is just 46.8m/s^1 multiplied by 15s\n= 702m << note this as well.\n\n\nWe have one last section now. Here, it is accelerating at -2.9m/s^2 from a velocity of 46.8m/s^1 (the constant velocity we calculated remember) to a velocity of 0m/s^1 (rest). So we need one of the equations of motion here.\n\nWe know u=46.8m/s^1, a= -2.9m/s^2 and v must = 0m/s^2 \nwe want s. No t is involved here so the equation that doesn't have t is \nv^2=u^2 + 2as\n\nThis equation isn't giving us displacement in this form though. We need to rearrange to make s the subject by leaving it alone on one side of the equal sign.\n\nFirst lets put u^2 to the other side of the equal sign (iam taking it away from both sides of the equal sign to do this). \n\nThat gives us v^2 - u^2=2as\nNow we just need to divide both sides by 2a to leave s on its own.\nThat gives us \n\n(v^2 - u^2)/2a = s\n\nNow referring back to our values, we just subsitute them in,\nwhich gives us\n\n(0^2 - 46.8^2) / (2 x -2.9) = 377.6275862m << note this lastly\n\nRemember all them displacements that we noted, now all we have to do now is add them together. To find the total displacement travelled. That is \n\n421.2m + 702m + 377.6275862m = 1500.827586m\nWhich is your answer\n\nA few notes here:\nThe equations we used gives us displacement only, not distance travelled. That's why I always referred to them as displacements. And same for velocity. The equations only deal with velocity, not speed. The difference is that speed doesn't care about it's direction, velocity does. However, the automobile just travels in a straight line and straight forward in this case; there isn't any case where it travels backwards or anything. It just travels in one direction, so displacement automatically gives us distance travelled anyway and velocities in this case is the same for speed.\nAlso I didn't round up the any numbers (especially that long 377.62... number) becuase I don't want to round numbers until I get to the final answer, becuase that will introduce inaccuracies. I can however round up now at the final answer. \nI leave tha
908729
(a) ω = ωo + αt\n0 = 150 + α * 132\nα = -150/132\n≈ -1.136 rev/s²\n\n(b) θ = ωot + ½αt²\n= 150 *132 - ½*1.136*132²\n≈ 9903 revolutions (nearest rev)\n\n(c) See (b)\n\n(d) 75 rpm = 75*2π rad/min\nα = rω²\n= 50 *(75*2π)²\n≈ 11103300 cm/min²\n= 111033 m/min²\n= 111033/3600 m/s²\n≈ 30.84 m/s²\n\n(e) ???? I do not understand what it is you are seeking.
210676
Use free body diagram to balance the forces and u can derive the equation of motion \nIf u want i can give u full derivation.\n"::
927736
Everyone above assumed a constant speed.  the more general formula is \n\nD = ∫v dt
864633
He's right. You take the derivative of the equation and plug in the values for t (just consult any book on differential calculus for details on this)\n\nSo h'(t)= -3.92t+10 (the derivative)\n\nWhen t=2.3, the velocity will then be (-3.92)(2.3)+10, or .984 m/s\n\nWhen t=2.7, the velocity will then be (-3.92)(2.7)+10, or -.584 m/s (.584 m/s in the opposite direction)\n\nNot bad for a high school sophomore who taught himself differential calculus, eh?
856026
Use the gas law\nPV=nRT
769961
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