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?
211094
62% of 5.4 X 10^3 = 5400 * 0.62 = 3348 Joules
468149
a)F = MV=  m1v1 +/-  m2v2  \nusing these we get,  95*4.1 + 85*5.5 (since same direction)=857kgm/s = (95+85)* V. so, V = 4.76m/s\nb) MV = M (6) = (9500+m)*6 = 9500*16 + m*0. \nso, (9500+m) = 9500*16/6= 25333.33 kg\nso, m = 25333.33-9500 =15833.33kg\nc)work done against friction= kinetic energy of bullet\nW = FScos(tetha)  since its horizontal, angle is zero, \nand cos0 = 1, so W = FS\nwe know, coeff of friction = tangent of angle between, the frictional force  and normal reaction...\nso, the \nF = normal reaction=(1.1 +0.015)* 9.81=10.938N\nfrictional force= 0.25 *10.938 = 2.7345N\ndistance is 9.5m...\nso W = 2.7345*9.5 = 25.97 J\nthe kinetic energy of bullet has been transferred to the block...\nso the kinetic energy of bullet is 25.97J\n1/2*(0.015)* v^2 = 25.97\nso, v^2 = 25.97*2/0.015=3463.74\nso, v = 58.85m/s\nhope this helps...:)
501069
1. KE = (1/2)*m*v^2 - you have the KE and mass.\n\n2. The amount of work done will be equal to the kinetic energy the arrow is given. \nW = F*d = (1/2)*m*v^2 solve for v.\n\n3. When the speed is increased 50% Vnew = 1.5*Vold. Kinetic energy will increase by 1.5^2 = 2.25. \n\nWork done stopping it is Fbrakes*distance and equals the kinetic energy it has. \nKEnew = 2.25*KEold so the stopping distance, SD, changes accordingly:\nSDnew = 2.25*SDold\n\n4. The acceleration on the baseball is given by\nV^2 = Vo^2 + 2*a*d where V is final velocity, Vo is original velocity, a is acceleration, and d is distance. Finally, \nF = m*a.\n\n5. \n(a) The tension is divided between holding the weight and accelerating it. T = W + m*a.\n(b) The work gives the load a potential energy. \nU = m*g*y\nIt also gives it kinetic energy. The velocity it has at the 21.0 m height is given by\nV^2 = Vo^2 + 2*a*y. Use the velocity in the KE equation. The work done is the sum U + KE.\n(c) Might be a trick, but this feels like semantics. I don't see that \nb & c are any different. Maybe in class, or the book, some distinction was made between those 2 ways to say it?
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
174304
Your looking for the time they work together so let (x) the time they work together .the rate should be A=15,B=5 then the work should be A=x/15 and B=x/5 since your looking for the time...solve for it cause I'm not giving you all the answers. it should look like this \n\nx/15+x/5= the time they work together...\n\n(Next time listen to your teachers)
446702
(a) Gravity exerts a downward force on the block of 9.8 Newtons/kg * (0.25 kg) = 2.45 Newtons. Work done by gravity = force * distance = 2.45 Newtons * 0.12 m = 0.294 Joules\n\n(b) The spring is a bit tougher since the upward force increases as the spring is compressed. Force = 250 N * (C) where C is the distance the spring has been compressed (in meters). Work is then the integral of Force over distance, so Int(250 * C (dC)) from C = 0 to 0.12 M . The integral is 125 * C^2. Evaluation at 0 gives 0. Evaluation at 0.12 gives 125 (0.12)^2 =  1.8 Joules. So the work done by the spring is -1.8 Joules . Note that the work is negative, because the object is moving against the applied force.\n\n(c) When the block strikes the spring, it will have a kinetic energy of (1/2)mv^2 = (1/2)(0.25kg)v^2 = 0.125v^2.\nThis kinetic energy is absorbed by the spring as it compresses. Note that the work done by gravity is also absorbed by the spring.\nTherefore, we have the following equation:\n0.125v^2 + 0.294 = 1.8\nv^2 = 8(1.8-0.294) = 8(1.506) = 12.048\nSpeed of block just before it hits = SQRT(12.048) = 3.47 meters/sec downward.\n\n(d) I'll leave this as an exercise for you. Just apply the same techniques as above. Note that the block hits the spring with a velocity of 2(3.47) = 6.94 meters per second...
387156
Equate distances and solve for time.\n\ngraces distance: 5*10 + 5*t    \nwill's distance: 15*t\n\nNote that I've started the clock when will starts.  That's because the question asks how long will must ride.\n\nNow, just equate the distances and solve for t.\n\nI get 5 minutes.\n\ngrace = 50+25 = 75\nwill = 5*15 = 75\n\nBtw, I graphed these on my TI-83 and looked for the intersection.  It's helpful to draw the lines.
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
512161
You haven't given me any information with which to solve this problem.  How big are the blocks?  How much do the blocks weigh?  How far apart are they?  How much resistance is there when I push the first block?   How fast is the first block going when it is pushed?  Without all this I cannot figure out how much momentum is lost between the time the first block is pushed and when it hits the second block or if the second block will have enough momentum to move at all or make the third block move if it contacts it.\n\n\nBy the way, you cannot solve this without knowing how much each block weighs, how much momentum it has to begin with, how much friction there is (to reduce momentum), and how far each block has to travel before it contacts the next one.    You need to have the variables to create a formula. \n\nGeeze, don't you hate it when idiots try to act smart????
461787
Rate * Time = Distance\n\n2x = Rate (x) * Time (2) for first two hours\nx-10 = Rate (x-10) * Time (1) for third hour\n\n2x + (x-10) = 152\n3x-10 = 152\n3x =162\nx = 54\n\nThe correct answer is B. 54mph
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!
478997
#1:  20 miles.  The two cyclists are traveling a total of 10 mph, so it will take them exactly one hour to meet.  The fly is traveling at 20 mph, so he will fly 20 miles in that hour.\n\n#2:  12 cents.  Take one of the three-link pieces (let's call it group A), cut each of those three links (6 cents), then weld piece A1 between groups B and C (2 cents), A2 between C and D (2 cents), and A3 between D and E (2 cents). Voila!
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)
506598
If you run x1 meters against wind, the work you produce is A1=F*x1+f*x1; F is the force of your muscles to overcome the resistance of air with no wind, f is the force of the wind in your face. Let x1 be covered in time t1, then P1=A1/t1 is your power against the wind. \nIf you run x2 meters with the wind, the work you produce is A2=F*x2-f*x2; F & f being the same.\nLet x2 be covered in time t2, then P2=A2/t2 is your power with the wind. They say that dP=P1-P2=50W or dP=(F+f)*(x1/t1)-(F-f)*(x2/t2). As x1/t1=x2/t2=v, then dP=2*f*v. Thus f=dP/(2*v)=50/7=7.14N  is the force of the wind, m=80kg is excessive datum!
339201
Use conservation of impulse and conservation of kinetic energy.\n\nmv + 0 = mu + mw\n(1/2)mv2 = (1/2)mu2 + (1/2)mw2\n\nu + w = 1 (consider it as a straight line in a graph)\nu2 + w2 = 1 (consider it as a circle in that graph)\nWhen you solve this set of two equations, you get\nu = 0 and w = 1 (m/s)\n(Find the useful intersection. The other one is is the initial situation.)\n\nTh
106450
The efficiency of an ideal engine is (1- T2/T1) or (T1- T2)/T1 = (600 - 400) / 600 =0.333.\nBut in the problem it is given that it is 0.5 which is not possible.
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.