How to measure amout of force applied by a person? reply here email to me please denisecortez_09@yahoo.com?
951702
Weight, strength, and speed are the factors.◘
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.
446638
impuse is the change in momnetum which is mvfinal - mv initial\n\ntherefore 10*1 - (-25)*1 = 35 kg/ms. Took down as minus therefore 25ms becomes minus in the equation.\n\naverage force is therefore just the impuse divided by the time of contact and is therefore \n35/0.02 = 1750N
903133
From what I learned in physics class, the force you are speaking of is measured in Newtons.
368233
N is Newtons, a unit that measures force. For every 1kg of something, if it accelerates at 1 meter per second per second (m/s/s or meters per seconds squared), than there is a force of 1N acting on it.\n\nForce is calculated by multiplying the mass (not weight) with the acceleration. For example, if you are 75kg, than there is a force of 75kg X 9.8 m/s/s (acceleration due to gravity: roughly constant on Earth) = 735N pulling you down. BUT, there is a reaction force, in this case the NORMAL, pushing you back up at the same time and force. Therefore, the force pulling you down, 735N, take away the NORMAL, also 735N, and you will have a NET FORCE of 0N because you are stationary. \n\nSo in order to resist something pulling at 14N (ignoring air resistance, friction, etc), it needs to pull back 14N.
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
784877
How are they applied? Tangentially? In opposite direction? Or same?\nLet us assume oppo direc and on the border.\nThen net force=35N.\nTorque = r x F = I w\n=35x0.341= ''0.5'' m R^2\nR radius m mass\n ''Check if it is 0.5 only for a solid disk or cyliner about its axis.\n\nHope it helps
871072
since its you science homework the answer your more than likely looking for is Newton meter, now its up to you to find out what you can on it
4309
The force exerted by an orthodontic band can be measured by first calculating its force-constant.  It can be done in the following way:\n\n(1) Hang a single, unstretched, orthodontic band from a string.\n(2) Along side the band, place a measuring ruler and carefully note and record where the bottom part of the band starts.  This will be our starting point.\n(3) Next, hang a couple of known weights such as paperclips, fishing sinkers, etc, and note how far the band stretches with the given weight attached.  Perform this task for, say, 5 to 10 different weights.  Don't forget to measure and record how far the band stretched for each weight you hang from it.\n(4) Now, using graph paper, make an x-y plot of the data.  Mark how much weight was used (y-axis) and how far the band stretched (x-axis).\n(5) Through these data points, a straight line can be drawn which has a calculated slope.  This value of the slope, k, is your force-constant.  Don't worry, this will make sense in a minute.\n\nTo calculate the force exerted by the band, use the following equation: F = kx, where F is the exerted (restoring) force, x is the distance the band stretched, and k is the force-constant we calculated.\n\nViola!  There you go.\n\nDepending on the material, the force-constant k will be different.  This was why we had to go through steps (1) to (5).
853308
The net force on the man is the sum of the force of gravity and the force on the ground. You've already calculated the net force, so simply subtract the man's weight from it. Remember that gravity is pulling him down, so you'll be subtracting a negative number (i.e. the final result should be greater, not less, than the net force).
912139
The abverage acceleration is a = (V^2)/2x = 35^2/(2*.11) = 5568.18 m/s^2\nMultiply this by the mass and you get the average force:\n\nFa = .14*5568.18 = 779.14 N
72896
Newtons is the unit for measure of force
943484
I expect that you have the data on stretch as a result of tension. \na) The object is just at rest. Tension, T = W = m*g.\n\nb) T = W + m*a\n\nc) Tmax = W + m*a
814690
The forces on the person are\n\nSFx = 0 = N - Fc\nSFy = 0 = Fs - m*g\n\nwhere SFx and SFy are the sums of the forces in the x- and y-directions, respectively, N is the normal (to the wall) force, Fc = m*v^2/r is the centripetal force, Fs = u*N is the friction force, m is the mass, and g is the gravitational acceleration.\n\nThese equations say that\n\nN = Fc = m*v^2/r = m*w^2*r\nu*N = u*m*w^2*r = m*g\n\nwhere w = v/r is the angular speed. Solving for the coefficient of static friction gives\n\nu = g/(r*w^2)
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Okay,\n\nQuick method:\n\nShe picks up a certain K.E. (kinetic energy) before she hits the water and loses this K.E. 3m under. \n\nWe have KE = Force x distance = mad\n\nso we have m x a1 x d1 = m x a2 x d2\n\nm =65kg, a1= 9.8m/sec, d1=13m, d2= 3m\n\nF(ave)= m x a2= m x a1 x d1/d2 = 2760 1/3 N \n\nThe force exerted on her by the water from when she hits until she stops 3m under will not be constant but the average force stops her in a distance of 3m and has the above value
652894
the amount of pressure exerted on an object
555699
I think I can help you with this one... I learned how to do it in Kindergarten.\n\nPull your head out of your ass and measure.\n\nGood Luck!