an elevator and its load weigh 160N.?
461030
For the first problem, you need a law of motion. (final velocity)^2=(initial velocity)^2+2*acceleration*displacement. Substituting, you get 0=30^2+2*a*20m and a=22.5 m/s^2. Now, we need to write a force equation: ma=Tension force-weight. To find the mass of the elevator and its load, divide its weight by gravity (9.8 m/s^2) to get about 16.3 kg. And you already calculated the net acceleration. Our force equation can now be rewritten as (16.3 kg)*(22.5 m/s^2)=Tension force-160N. The tension force is about 207N.\n\nFor each of the weight scenarios in the second problem, you know the net acceleration to be 30 m/s^2. You can also find the mass of the passenger by dividing his/her weight by gravity. With the force equation ma=(force of feet up)-(weight) with respect to the passenger, you can find the force of his/her feet. Here is the work for the first scenario.\n\n16N/(9.8 m/s^2)=1.63 kg\n1.63 kg*30 m/s^2=(force of feet)-16N\nForce of feet=64.9N\n\nAnswers for the second and third scenario:\nb) 2600N\nc) 260N
229380
It will weigh things that are suspended in air or water. It was invented by Newton.
367999
It weighs the amount that the air inside it weighs.
524636
If the forces are both acting in the same direction, add them together. = 65 Newtons
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?
1052952
1650 lumber has a specific gravity of about 0.42.  This means its density should be close to 26.19 lbs/ cu. ft.  The one foot 2x4 should weigh 1.46 lbs.
353265
Hess's law \n\n\n +142KJ  that is all you want "right" is the answer
965597
you convert mass into force by multiplying with 9.8m/sec^2 which is the acceleration due to gravity
880761
People above neglected the vertical component of 460 N.  This must be added to the weight of the box.  Also, not sure why they took 350 N and multiplied it by acceleration.\n\nso F_vert=460sin35.6=267.78 N (downward)\n     F_hor=460cos35.6=374.0 N\n\ncalculating F_frict=mu x (F_normal)\n\nF_normal=weight of box + F_vert = 267.78+350=617.78\n\nso F_frict=0.57x617.78=352.13 N\n\nnow the net force on the box = 374.0 - 352.13 = 21.87 N\n\nand we need mass of the box, m=350/9.81=35.7 kg\n\nso a=21.87/35.7=0.6126 m/s2\n\nfinally,  d=V_initial x t + 0.5 x a x t^2, V_initial=0\n\nso t=((4.2)/(0.5x0.6126))^0.5=3.7 sec\n\n**********************************************************\n\nHowever, should you wish to ignore the vert component\n\nF_frict=mu x (F_normal)=0.57 x 350 = 199.5 N\n\nso net horizontal force is:\n\nF= 374.0 - 199.5 = 174.5 N, a=174.5/35.7= 4.89 m/s2\n\nplug a into:\n\nd=V_initial x t + 0.5 x a x t^2, and solve for t\n\nt=((4.2)/(0.5x4.89))^0.5 = 1.31 sec
930053
The force of static friction is the product of the force normal to the surface and the coefficient of friction. This force must be greater than or equal to the passenger's weight. First, we find the acceleration:\n\na = v²/r = ω²r, where ω is the angular velocity in rad/s\nω=.5 rev/s = π rad/s\na=4.6π² m/s² ≈ 45.40018 m/s²\n\nThe normal force producing this acceleration is:\nF = ma = m*4.6π² m/s²\nSo the maximum force of static friction is:\nμ*m*4.6π² m/s²\n\n(Note: those units are correct - I'm not writing the kg because the units of mass are contained in the term for mass - if I were to substitute a particular mass, I would get something like μ*(60 kg) * 4.6π² m/s², so writing kg a second time would be redundant).\n\nNow, the weight of a passenger is:\nF=mg ≈ m*9.81 m/s²\n\nSo we have:\nμ*m*4.6π² m/s² ≥ m*9.81 m/s²\n\ndividing by m*4.6π² m/s:\n\nμ ≥ 0.216079\n\nHowever, since our initial values were only given to two significant digits:\n\nμ ≥ 0.22
173724
If it is a fixed pulley, you need the same force as the weight of the load which is 9.8*0.2 Newtons (9.8 is the acc. due to gravity and 0.2kg is the mass of the load) or simply 200gf (where gf is gram force). Now you can tell one half of this number.\n\nAlternately, if you have a movable pulley, the force required is half the weight of the load, i.e., 100gf.  You can tell one half of this number too.\n\nIf you need to find the difference between fixed and movable pulleys, look up a physics text (or science text), I am not totally familiar with the schooling system here.   Otherwise, Google should show you pictures of fixed and movable pulleys.
456850
Do your home work. \n\nStill I tempt to answer this question.\n\nW= weight of the box\nmu= coefficient of friction.\n\nWe assume there is no acceleration. So force to upward movement is Fup = Wcos(20)*mu + Wsin(20),\n\nsimilarly Fdown = Wcos(20)*mu - wsin(20),\n\nso\nWcos(20)*mu + wsin(20) = 4Wcos(20)*mu - 4wsin(20)\n5Wsin(20) = 3Wcos(20)*mu\nmu = 5sin(20)/3cos(20)\nmu= 5/3 tan (20)\nmu = 0.606
3541
all you have to do is calculate the work that the engine should have done.  W=Fx meaning work equals force times distance.  The force is mass times acceleration, so it would be 10 x 9.8 (9.8 meters per second squared is the acceleration due to gravity) giving you 98 (lets round to 100 so i dont have to get a calculator or think too much because im tired, and your other data only has one significant figure anyway, if you think you need it more precise you can go get a calculator and do the work im going to show you with 98 on a calculator...)\n\nanyway, so to find the work (W=Fx) you multiply the force we just found (100N) by the distance so you get 100N x 4m = 400J.\n\nNow you know that the engine should have done 400J of work, but instead it did 500J, meaning that it did extra work, so it's not 100% efficient, so to find the percentage of its efficiency, just divide 400/500 and multiply that by 100 to get the percentage.  So its .8 x 100 so 80%.\n\nAnd if i may ask, why do you keep asking very basic questions on completely random physics topics that you would clearly not be doing all at once, and if you already did all of it, you should already know these answers?
306356
You would be hurt more.  If you are in contact with the elevator, It would cushion and defray some of the impact force.  If you are in the air, it's just you and the ground......
355325
Newtons, or sometimes kilograms
952952
one third of the elastic limit is 8x10^7 Pa and the total area is 3x10^(-4) m² so the total force is 2.4x10^4 N.  It takes 12,740 N to hold the elevator stationary agains the acceleration of gravity which leaves 11,260 N available to accelerate the elevator upwards, so the maximum acceleration would be about 8.66 m/s².\n\n\nDoug
718152
weight equals mass times acceleration due to gravity