a certain bicycle has an efficiency of 62 percent.  how much work is done by the bicycle is the cyclist does 5.4 x 10 to the 3rd power J of work on the bicycle?\n\nthe equation to use is  v=F multiplied by the upside down y\n\nhow do i do this?
126073
You are looking for the amount of work done by the bicycle.\nWe know this value will be less than 5.4X10^3 J because energy is always lost when using a machine.\nIn addition, the answer is equal to the efficiency (62%) times the amount of work the cyclist does. \n(I wouldn't be right if I did your whole homework problem so I'll stop here)
1012
A. Potential energy = Force * Distance\n     = (50 * 9.8) * 10\n     = 4900 J\nB. Power rating = Energy / time = 4900 / 5 = 980 Watts
420411
Part a)\n\nVf^2 = Vi^2 + 2 * a * d\na = (Vf^2 - Vi^2) / (2 * d) = -4,489/162 = -26.7 m/s^2\nm = 220*10^3 / 9.81 = 22,426 kg\nF = m*a = -598,777 N\n\nW = F * d = 50,297,245 J\n\nPart b)\n\nF (from above) = 598,777 N\n\nI hope this helps!
406704
It usually helps with these kinds of problems if you start by drawing a little sketch. It doesn't have to be accurately scaled, its purpose is to help you recognize the physical relationships and to keep your signs correct.\n\nYou are dealing here with rotational motion. You should be familiar with the relation: (alpha) * J = T, where (alpha) means the rotational acceleration, J is the rotational moment of inertia, and T is applied net torque.\n\nYour bicycle wheel has inertia. In general the rotational inertia of an object is (integral)[r^2 dm] where r is measured with respect to the axis of rotation. That's very simple in your case since all your mass is distributed in a thin ring - so r is constant, and you get J = r^2 m. This works out to 0.1941 kg m^2.for your problem.\n\nTorque due to a force applied at a distance from the center of rotation is T = r * F. Now here's where the sketch starts to become useful. The problem tells you that you have one force (123 N) applied at the rim and a second force applied to a sprocket 8.92 cm in diameter, and the net result is acceleration of 4.49 rad/s^2. And the problem tells you that the 123 N force is resisting the motion. So you get:\n\n(alpha) J = T1 + T2\nwhere\nT1 = what we seek\nT2 = -123 N * .3205  m  The (-) sign is because it opposes T1.\n\nSolve this to find T1 = 39.6 N-m\ntherefore the force required is T1/r1 = 39.6 / .0446 = 888 N.\n\nIf you have followed this you should be able to do the second part easily.
87941
Average Power = Work/Time\nWork=Force*Distance\nDistance=2*pi*radius=2*pi*(0.31 m)=(1.95 m)\nAverage Power = ((25 N)*(1.95 m))/(1.6 s)=30.4 W
37166
That's a tough 8th grade question; I've seen similar ones in algebra 2.  You have to figure the amount of the job that gets done in one hour. If x is the total time for the faster machine, 2x is the total time for the slower machine.  Together it takes 3 hours, so together they can do 1/3 of the job in one hour.  The slow machine can do 1/x of the job in an hour; the faster, 1/(2x).  This gives the equation:\n\n1/x + 1/(2x) = 1/3\n\nmultiply both sides of this equation by the LCD (6x) as if it were a giant distributive property example and get:\n\n6 + 3 = 2x       (the denominators all cancel out)\n\nso 9 = 2x (the slower machine takes 9 hrs)\n4.5 = x (the faster one takes 4 1/2 hours)
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
383773
Well first just ignore the 2.425 part of the equation, because it does not depend on X.  \n\nI went by trial and error, here are some (X, C(X))'s...\n(1, 1.825), (2, 1.425), (3, 1.225), (4, 1.225), (5, 1.425), (3.5, 1.2)\n\nSo the answer is 350 bicycles.\n\nSorry I can't show more work.  I assume you have to factor it into a form of (Ax + B)(Cx - D).
457402
Come on now, what is the formula for this???  :)\n\n(V1 * T1)/P1  =  (V2 *T2)/P2\n\n(69L * 131)/2.85  = (xL * -37)/2.85\n\nNow, solve for 'x'?  Be sure to watch your units!!!!
536645
This is an energy balance situation.  The work done in dragging the object must equal the energy change.  In this case the energy is potential energy, given by m*g*∆h, where ∆h is the change in vertical distance.  The work (energy) required to drag the block is force times distance along the incline.  Equate these two quantities and solve for F.
86998
work = (200N)*(10.0m)*cos(0) + (50.0N)*(35m.0m)*cos(0)\nwork = 2000 Joule + 1750 Joule\nwork = 3750 Joule
430265
1) 4.2cycles/41.4s = X/1s\nX=0.101Hz\n\n2)40cycles/84s = X/1s\nX=0.476Hz
237582
OK, the barrel´s mass is 40Kg (notice: its mass is 40 Kg, not its weight). \n\na) So, your formula is Weight = mass(m)*gravity´s acceleration(g)= 40Kg*10m/s^2 = 400 KG*m/s^2 (=400 Newton)\n\nb) Work = force * distance\nSince work is calculated only for the vertical, and the force equals the weight, you have:\nWork = 400N*5m = 2000 Joule\n\nc) Eff = 250N*5m/2000J*100 = 0.625*100 = 62.5%\n(You got 12.5% because you forgot to multiply 250N for 5 meters. Remember that efficiency is a relation of works, and work is force * DISTANCE).\n;o)
434805
The energy absorbed is that which is not outputted.  In this case, 75% of the energy is absorbed, so the energy absorbed per cycle is .75*8500 J.  One watt = 1J/sec.  So seconds = J/watt.  The power is constant, so the power in each cycle is 6,00 kW; this times the duration of the cycle will be the energy in each cycle.
458485
1)  Notice that 6^2=6*6=36; 6^3=36*6=216; everytime you multiple a power of 6 by 6, the units digit remains 6.  Now what about powers of 5?\n\n2)  If r is total minutes and the first 3 minutes is $1.00, and the rest of r costs $0.20 per minute, then the total cost would be $1.00 + $0.20(r-3) = \n1 +0.2r + 0.6 =\n1.6 + 0.2r\n\n3)  3*10^4 = 3*10,000 = 30,000, and 4*103 = 412.\nNow divide 30,000 by 412 to get their ratio, and then multply that number by 100 to convert to percent difference.\n\n4)  Not enough info - you need the height too.  If the same height, obviously the can with the larger radius\n\n5.1)  65*20% =65*0.2 = 13\n\n5.2)  Again, not enough info. From the info given, all you can say about one-half the number of people who preferred green was that it was between 0 and (65-13)/2, aka 26
69666
Output work 3.2J\nInput work 12 x 0.3 Nm = 3.6 J \nEfficiency = output work / input work = 3.2/3.6=0.88=88%
132613
well, using f=ma, and w=fd, we immediately get w=(ma)d\nW=work  (3650 joules)\nM=mass  257kg\na=accelleration =gravity =9.81\nd=distance (required)\n\nrearrange w=(ma)d\n\nd=w/(ma)\nDistance=work/(mass*acceleration)\n\n= 3650/(253*9.81)\n=1.47m