How much work does a man do in holding10Kg mass at a height of 5m for 4hrs?
952970
Work is force times distance.  Since he is not moving the mass, he is doing no work.  \n\nHowever, his body has used a lot of energy and his muscles are tired.  This is because your body needs energy to stay upright and balanced.  It needs even more energy if you are balancing a load.  Unfortunately, all that energy has been converted to heat, and not towards doing any work.  What a waste.
787792
If Mr. Thomas practiced 5 days I have to assume that he practiced 1hr 5min, and 45min each day...right?  If so then first find out how much that equals each day.  1hr 5min + 45 min = 1hr 50 min.  Now how much is that for 5 days? Separate them. 1x5=5 and 50x5=250. 5hrs 250 min...ok...almost there.  How long is 250 min? Let's turn it into hours and min...250 divided by 60 (because there are 60 min in an hour) is 4 hrs with a remainder of 10. So add the first 5 hrs with this 4 hrs...9hrs and tack on the remainder of 10 min....9hrs 10min.\n\nMr. Welch worked 15 hrs and 45 min more last week.  change 40 hrs to 39 hours and 60 min....they are equal as there are 60 min in an hour so 39 hrs 60 min minus 24 hrs 15 min...39-24=15 and 60-15=45....15hrs 45min.\n\nI think there are 16 ounces in a pound, right? if so then the five babies' average weight is 6 pounds 8 ounces.  Start by turning the pounds into ounces...32x16=512  then add the other 8 ounces...512+8=520...now divide by 5   520/5=104...ok now we know that the average weight of each baby is 104 ounces.  Let's turn that back into pounds and ounces...divide 104 by 16 (16 ouces per pound)  equals 6 pounds with a remainder of 8.\n\nLance is 6.75 feet tall.  12 inches per foot so we have to determine how much 9 inches of one foot is.  9 divided by 12 is .75 or 3/4.  so I don't know how your answer needs to look but the answer could be 6.75 ft or 6 3/4ft.\n\nDenise traveled 804.5 kilometers.  One mile equals 1.609 kilometers.  So 500x1.609=804.5.  look in your book and see what conversion factor they give you and make sure that it is 1.609...they might call it 1.61 or 1.609344 or something close, but regardless, just multiply 500 by whatever that number is.\n\n\nWhew...that was a lot of work!
241491
probably around 90kg.
876805
Work = F * d\nWork = Force of the spring * distance mass moves\n\nWork refers to an activity involving a force and movement in the directon of the force. A force of 20 newtons pushing an object 5 meters in the direction of the force does 100 joules of work.\n\nThe max KE is the work put into moving the object.
514815
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)
730564
1. friction\n2. W=mg  , he has a mass of 100Kg. Weight is the force=newtons, g=pull of gravity N=100 x 9.8m/sec^\n98Newtons\n3. correct
80396
Yeah i know:\nDo this:\n\na.)Find the work done by the applied force.\n\nb.)Find the work done by friction\n\nc.)Find the work done by gravity\n\nd.)Find the work done by the normal force.\n\ne.) IF the object starts at rest, how fast is it going after the 5.0m.\n\nhope this helped\n-----------\n\nNO I WILL NOT HLEP YOU. DO YOUR OWN WORK! your asking someone to do your work for you.
701
I work out 5 to 6 hours plus my work or job is a real work out.
153995
F=m*a\na=F/m = 3.5 x 10^7N/(2.0x10^6kg) = 17.5 m/s^2 or 1.8g
131445
Torque is force*distance.\n\nFind the weight of the person (M*g) = 89kg *9.81m/s^2 \nM*g= 873.09 Newtons\n\nthere is a 1:3 lever arm there--> 4cm/12cm\nThe muscle is having to exert a force of 3*873.09N or 2619N\n\nThe torque around the pivot is 2619N @ 4cm (or .04m)\n2619N * .04m = 104.77N*m\n\nYour question says you want to "find the torque" yet you ask for the "forces pushing up".  Make sure you know which one you need.
969410
Given: m=1.6*10^7kg\n           F=8.5*10^5N\n\nI only guessed that the above is what you meant.\n\nSolve for a, using the formula:\n\nF=Ma\n\n8.5*10^5=1.6*10^7a\na=8.5*10^5/1.6*10^7\n  =5.31*10^-2m/s^2\n\nv=70km/h =70*1000/3600m/s\n  =19.44m/s\nu=0\n\nTo solve for time, t, use the formula:\n\nv=u+at\n\n19.44=0+5.31*10^-2t\nt=19.44/5.31*10^-2\n =3.66*10^2\n =366s
951151
Weight the person has to lift = mass of the bucket * accelaration due to gravvity\nWeight the person has to lift = 24 * 9.8 J \nWork done = weight lifted * DIstace\n8*1000 J = 24 * 9.8 * Distance\nDistance = 24*9.8/8*1000 meters\nDistance = 3*9.8/1000 meters\nWell this is the answer. Right now i am feeling quite lazy to calculate all this. Hope you can manage with this answer. \nThanks
52451
10 Hours\n10*7 = 70 Hours \n10*7*4 = 280 Hours
316987
6 per 10 seconds = 36 per one minute.\nTherefore\n36 (Push-ups) * 60 (Minutes) = 2,160\n\nBut it is impossible to do 2,160 push-up in an hour, IMO.\nGood luck though, if it works for you I will definelty try it out.\n\nAnd if not at least post how many you could make.
615213
If you are 5'4", you start with 100 pounds for the first five feet.  Then you add five pounds for every inch over five feet.  If you have large bones, you add ten pounds to the total.  Thus your weight would be 120 to 130 pounds for the right weight.\n\nIf you are a man, you start with 110 pounds for the first five feet and add 5.5 pounds for each inch over five feet.  Thus a 5'10" man should weigh 165 to 175 pounds.\n\nThose were the weights before the world started overeating.
577696
Assume the man is standing b met res from the aerial. \nso, b=100*tan49=115m\nheight of the top part aerial=b/tan52\n=115/tan52=89.9=90m\nThus, the total height of aerial = 100 + 90 \n=190m (roughly).
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)