problem of the week #1 how much dose each bale weigh?
144522
i don't know but is it actually some kinda of a question coz i just didn't understand it
348023
weigh 3 in one and 3 in the other:\n\nif they are equal:\nweigh each of the other 2 in one pan each. One will be lighter, voila\n\nIf they are different:\ntake the lighter set of 3, weigh one in one pan and one in the other and hold one aside.  If one is lighter, voila, if they are equal it's the one set aside.\n\nthat's the whole thing in 2 weighings
257076
dude u need to start doin your own hw. ill do it this time\n5. well you have to check all the forces on the bag. the first force is gravity, which is downwards and its 9.8m where m is the mass (Force = mass x acceleration) so gravity force is 20.5kg * 9.8 = 200.9N. then the 2nd force is the force of your own movement. you accelerate it upwards at 5m/s^2. so you apply a force of 5x20.5 = 102.5N on the bag upwards. but this is the thing. it gets confusing. the force is applied on the bag, not on the items in the bag. the items in the bag  is what  makes the bag weigh so much. so if you pull the bag up with 102.5N force, the items in the bag pull downwards with the same force (another one of newtons laws) which means the total downwards force is 102.5 + 200.9 = 303.4N downwards. which is more than 245N so the bag will rip. \n\n6. well in general F  = ma. but the F is total force. in this case, there is a friction force along with the applied force.\nso Fa + Ff = ma\n   37 + Ff = 5 x 5.8\n   Ff = 29 - 37\n   Ff = -8N     its negative because its opposing the applied force. in this case i can choose which direction i want to be positive, i chose the direction that the object is moving. you could put negatives in front of the applied force, friction force and acceleration   and youll get the same answer.  so friction force is 8N.\nb. the coefficient of friction is calculated by Ff = u x Fn\nwhere Fn is the normal force and u is the coefficient. you already know Ff. the normal force is easy its basically  the force tha topposes gravity. if the object isnt moving up or down (if its on a surface). so then gravity is 5 kg x 9.8 = 49N \nso then Ff = u x 49\n    8 = u x 49\n u = 8/49\n u = 0.1632..... coefficient of friction is 0.1632\n\n7.\na.  the force is 107N and the mass is 5.1kg. so acceleration is 20.98m/s^2\nb. so acceleration = Vf-Vi/t\n  20.98 = Vf - 0 / 10s\n  Vf = 10 x 20.98 = 209.8m/s\nfinal velocity is 209.8m/s\nc. the net force that acts on the object is only the downward gravity force which is 9.8x 5.1 = 49.98.\nd. the direction of their velocity goes down when the velocity becomes 0. so you can do this by saying acceleration is gravitational and it is -9.8. initial velocity is 209.8 and final velocity is 0. and now u gotta find time. so  a = Vf-Vi/t\n-9.8 = -209.8 / t\nt = 209.8/9.8\nt = 21.4s   so it starts going down   21.4s after 10s after its been released (the first 10 s  its accelerating upwards)   so it starts going down 31.4s after its been realeased.
319440
Answer: 1,3,9 and 27 kg\n\nWay of solving:\n\n1) 4 pieces are very small number to weigh such number (40 of different weights. So I just tried to understand a most efficient way of weighing different weights using a small number of pieces.\n\n2) Firstly, I understand the most efficient set of pieces when we put some thing we are weighing on one scale and pieces on another. There are 15 (2^4-1) ways of taking different non-empty subset from set of 4 pieces. So, in most effective set all those subsets must have different weights. Eady to see, that pieces which are powers of 2: 1,2,4,8 kg make such a set (15 different weighings from 1 to 15 kg). \n\n3) Next I guessed that it is possible to put pieces not only on another but also at the same scale as we put some thing we are weighing. So, the weight of little piece can be not only added to the big piece, but also subtract from it, giving more possible combinations of weights.\nThen I guessed that some subset1 without smallest piece (ss1), ss1 + smallest piece (ss1+), another subset without smallest piece (ss2), ss2 with smallest piece on another scale (ss2-) must be consecutive numbers. So, ss1 = ss2 + 3. So, if smallest piece is x kg, the next piece must be 3 * 1st=3x kg.\nThe same situation is with other pieces: 3rd = 3 * 2nd = 9x kg, 4th = 3* 3rd = 27x kg.\nThis set of pieces makes possible 40 different weights.\nAlso, we know that sum of weights of those pieces is 40 kg.\nSo, x + 3x + 9x + 27x = 40kg\nx = 1kg\n\nAnd the final answer is: 1 kg, 3 kg, 9 kg and 27 kg.
101463
that depends on the material's specific gravity\nweight in kg = specific gravity * volume in ltrs
588958
the weight will be according 2 density of the material \n\nfor example for water 1  liter wighs 1 kg
703243
set up 2 equations with 2 unknowns and solve\n\none equation will involve the total poundage\nx + y = 9\n\nthe other equation will  involve your  costs with oh say the Right Hand Side having 9 * .96 as its value\n\n.... i want you to think on your own so this is as far as i go :)
621321
If the density is 'd' kg/m^3 then the weight is '9.81d' Newton. [remember weight is force]
581911
You only need to perform ONE weighing.\n\nFirst: take 1 match from the first box, 2 matches from the 2nd box, 3 matches from the 3rd box, ..., and 10 matches from the last box.\n\nWeigh the 55 selected matches. If the matches are all 10 grams, the total weight will be 550 grams. However, this is not true because some of them weigh 9 grams. For each 9-gram match, the total weight will go down by 1 gram.\n\nSo if the total weight is 549 grams, then the first box has the 9-gram matches.\n\nIf the total weight is 548 grams, then the 2nd box has the 9-gram matches.\n\n...\n\nIf the total weight is 540 grams, then the 10th box has the 9-gram matches.\n\nThus: Given this setup, the nth box has the 9-gram matches where n = (550g - total weight)
168241
It varies between weights. Anywhere from 25-32, most likely.
720122
yes they say its about ever 7 pounds for each size great job keep up the good work hun
188870
a scale or balence that uses grams
617024
There is no specific weight...it is your choice BUT the lighter the better!
429450
try some small branches. just stick them through the corners. \nif that doesnt work, wrap the top and bottom around the bale with string. \nisnt your bale tied? stick the target underneath the string\n\nor find someone and say "Here, hold this for a moment, yah?"
712680
You can "guesstimate" it weighs about the same as a cubic foot of water, since most prepared food products are mostly water anyway. So, about 62Lbs, or very roughly 28 kilograms...\n\nHope that helps...
843506
One pound is 0.454 kg\n100 lbs = 45.4 kg (mass)\n\n1 kg = 9,8 N (weight)\n45.4kg = 444.92 N
503781
http://www.weightchart.org/\ncheck this out