How much power is needed to do 1000 joules of work on a box if it takes 50 s to lift to it?
346550
kingbodonti is correct...\n\nPower in watts = Work in joules / time in seconds\n\nPower = 1,000 joules/50 seconds\n\nPower = 20 watts ANSWER
103276
The work done to lift the box to that height is stored in the body as the potential energy. Hence the work done is 400 J.\n\nAs for the height, potential energy is given by m*g*h, and the weight is given by m*g (=200N here).\nTherefore, 200h = 400 => h = 2m.
818883
Es = L*r*w (w=2piN) , N=600 rpm=10rps , L=9 \nEs = L*r*2pi*N) , N=10, L=9 \nEs = 9*0.18*2*3.14*10\nEs (Shaft Energy)= 101.734 Newton/second which is not a unit of power.\n\nPlease specify the units of r,Es,L=95 or 9 ?\nWhat is the unit of Ee? HP or KW?\nfurther you have given following data which are contradictory.\nPump Shaft Speed =1200 ?\nEe (Electrical Energy) = 67.2\nT = 1.62\nw (Rotational Speed)= 62.8
959600
need to know the mechanics of the lift you are proposing. Otherwise the answer is just 826 N.\n\nAre you worrying about compresive load? Tensile Load?  Which Joint?
508025
Dry air has a density of 1.168 kilograms per cubic meter.\nDry air has a heat capacity of 1.005 joules per Kilogram.\nSo it takes 1.17284 joules to raise the temperature of 1 cubic meter of air by 1 degree C.\nA Watt is a Joule per second, so it would take a 1 watt electric device 1.17 seconds to do the job.\nHumid air would require much more energy since it is denser and has a higher heat capacity.
827623
F*m = μ*Joules→ m = .25*40000000/1000 = 10000m = 10 km
487670
None. It takes no power to maintain that speed (on level ground, no friction). The object has kinetic energy (0.5 * m * v^2). Power was required to put it in motion and power can be recovered by decelerating it. If you do the math, the energy is 3551.86 ft-lb, equivalent to 1 horsepower for 6.458 seconds. Answer 1 neglected the fact that weight must be converted to mass by dividing by g, and energy is proportional to v^2.
367171
work=forcexdistance\npower=work/time\n\n1 in=2.54cm\n8x2.54=20.32cm\n\n10cm=1m\n20.32cm=2.032m\n\nwork=700Nx2.032m\n    =1422.4Nm\n\npower=1422.4Nm/10s\n     =142.24Nm/s is the power
302252
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)
963186
1000 Joule = 1kJ\n109.47J * 1000 = 109470\nSo, yes...it is.
783617
To give you a simple example...\n\nIf you want to move an object of 200kg it would require a force of 200kg.\n\nIf you use one pulley, it would require half of that force... 100kg\n\nIf you use a second pulley placed after the first one, you are reducing it by half again... and it requires a force of only 50kg\n\nUsing several pulleys is called "block and tackle".
661128
there is no thumb rule or some defined distance for that. all what you do is keep half meter distance and of-course it depends upon the KW of the power cable. and if you do not have enough space for keeping distance you can make a concrete slab in between
595586
I got 1650. First I multiplyed 825 * 5, and then divided that by the time which was 2.5. The formula for power if I remember correctly is work divided by time. Work is equal to f times d. D is the distance and f is the force in Newtons.\n\nI think this is correct if I remembered how to do it the right way. :]
612298
up to 10 trillion watts
520009
Another rule of thumb is to assume 100% load. \n\nNot knowing what kind of "electric motor calculations" you are going to make it is difficult to say what load factor you should use.
963007
multiply the mass of the water by the specific heat and the (temperature required minus the ambient temperature).that will give you the energy in joules.if you know in how much time you want this temperature to be achieved you can calculate the power as power is nothing but the rate at which the energy is dispensed
989843
net force IN THE DIRECTION OF DISPLACEMENTis (1.7*10^6 * cos 14)*2 \n= 3.3* 10 ^ 6 N\nwork done = 3.3*10^6*860=2.838 * 10^9 Joules