Units required for answer of   FORCE = MASS * ACCELERATION   ???
955176
Force - Newtons\nMass - Kilograms\nAcceleration - metres per second
723976
Question is: why on earth would you want to make that calculation?  m/s^2 is the standard form for acceleration.
966154
The magnitude of a force is measured in newtons(N)
894777
Kinesiology II\n Physics of Motion\nI.  Force \nA.  Definition\n1.  Force = mass * acceleration\n2.  Mass is a fundamental dimension, units g, kg\n3.  Acceleration is change in velocity, units m/sec/sec or m/s2\n4.  “Weight” is a force, units pounds or Newtons\n- Pounds = mass * acceleration of gravity\n- Why your weight, but not your mass, is different on the moon\n- Gravity on Earth = 9.8 m/sec2 \n5.  Example:  A person with a mass of 70 kg weighs:\n                        70*9.8 = 686 kg*m/sec2 or 686 N\n \nB.  Aspects\n1.  Force has a point of application or origin\nExample:  location where muscle attaches to bone\n2.  Force has a line of application\nFollows the tendon\n3.  A rotational force is called torque\n4.  Force has direction\nCan be indicated in two (x,y) or three (x,y,z) dimensional space\n5.  Force has magnitude\nMagnitude is measured in Newtons\n \nII.  Rotational Force \n1.  Force (mass*acceleration) may cause rotation about an axis\n2.  Moment of force (=Torque) is equal to the force times the lever arm – the perpendicular, straight line distance between the axis of rotation and the force\n3.  Moment of force (torque) = force* perpendicular distance \nUnits:  N*m\n4.  Sign convention:  \nClockwise moment = negative\nCounterclockwise moment = positive\n \nIII.  Levers – Location of effort force, load or resistance force, and rotational axis\nA.  Three types of levers\n1.  First class\n2.  Second class\n3.  Third class\n4.  Differences between the three have to do with the relative locations of the effort, load, and the axis of rotation\n \nB.  First Class Lever\n1.  “EOR”\nO = axis or pivot point\nE = effort force (as applied by muscle/tendon)\nR = resistance (load) force (outside force such as barbell)\n2.  The effort force and the load force are located on opposite sides of the axis of rotation\n3.  Example:  seesaw\n4.  Anatomical example:  Triceps muscle – elbow extensor\nInsertion:  Olecranon process of ulna\n5.  First class levers are rare in human body\n \nC.  Second Class Lever\n1.  “ORE”\n2.  The resistance (load) force and the effort force are on the same side of the axis of rotation\n3.  The effort lever arm is GREATER than the resistance lever arm\n4.  Example:  wheel barrow\n5.  Anatomical example:  brachioradialis – elbow flexor\nInsertion:  Styloid process of radius (distal – past the center of mass)\n6.  Second class levers are rare in the human body\n \nD.  Third Class Lever\n1.  “OER”\n2.  Effort and resistance (load) forces are on the same side of the axis of rotation\n3.  The effort lever arm is SMALLER than the resistance lever arm\n4.  Anatomical example:  Biceps brachii – elbow flexor\nInsertion:  Tuberosity of radius (proximal)\n5.  Anatomical example 2:  Quadriceps group – knee extensors\nInsertion:  Patella, tibia \n6.  Third class levers are common in the human body\n \nIV.  Mechanical Advantage\n     A.  Mechanical Advantage defined as:\nEffort Lever Arm/Resistance Lever Arm\n1.  A second class lever will always have a mechanical advantage greater than one because the effort lever arm is always GREATER than the resistance lever arm\n2.  A third class lever will always have a mechanical advantage less than one because the effort lever arm is always LESS than the resistance lever arm\n3.  A first class lever can have a mechanical advantage less than, equal to, OR greater than one, depending on the locations of the effort force and resistance force versus the axis of rotation\n \nB.  Mechanical Advantage VS. Speed\n1.  The mechanical advantage of a third class lever – common in the body – is poor – always less than one\n2.  However, the SPEED of rotation created by a third class lever is high\n \n\n3.  Because the origin of the resistance force is located farther from the axis than the origin of the effort force, it must travel a greater distance in the same time\n4.  Greater distance per unit time = greater speed\n5.  Good for throwing objects, kic
738706
Both questions use the formula F=ma (Force = Mass * Acceleration):\n\n1) F = ma\n1310 = 214*a\na = 1310/214 = 6.1215 m/s^2\n\n2) F = ma\n3.6 = m*9.8\nm = 3.6/9.8 = 0.367 kg
1036294
letter b is the answer.. because the weight of the object is approx. 627.20kg multiplied by 9.8(acceleration due to gravity) so the product is the force..
987931
you are making a mistake with the units\n\ndid you mean kg/s^2? if it is so the answer is 10300 kg/s^2\nif did you mean kgf/m it is 1050 kgf/m \nif not your units are wrong. \nremember that 1 N (newton) is 1 kgxm/s^2\nand 1 kgf = 9.8065 kgxm/s
621523
Newton's Second Law states that an applied force equals the rate of change of momentum. For constant mass: F=ma, or force equals mass times acceleration. In other words, the acceleration produced by a net force on an object is directly proportional to the magnitude of the net force and inversely proportional to the mass. In the MKS system of measurement, mass is given in kilograms, acceleration in meters per second squared, and force in newtons (named in his honor).
912301
Well, for centripetal force:\n\nForce = (mass) * (velocity squared) / radius\n\nand from there?  I'm lost.  I've tried plugging in all your information and it feels like we're missing a known variable.  My first assumption was to put in 7.18 as the force, but then we don' t the mass.   And if we plug in (7.17*mass), we're left with force as a variable.   \n\nOye, good luck.  It's been 10 years since I've taken physics.
680834
A force is a push or pull upon an object resulting from the object's interaction with another object.\n\nThe formula is F=ma, where F= force, m=mass, a=acceleration
962881
if acceleration is on the x axis, the slope represents 1/mass\n\nas\n\nslope = x/y = acc/force\nand\nforce=mass x acc\nso\nacc/force=1/mass\n\nif force is on the x axis, the slope represents mass\n\nas\n\nslope = x/y = force/acc\nand\nforce=mass x acc\nso\nforce/acc=mass
82803
Yes acceleration and force should have a linear correspondence in this experiment (noise excluded). Obviously Force = Mass * Acceleration.\n\nIt is difficult to give much more qualitatively without a better description though.
494540
Haha whats with all these answers, if you know basic physics, One of Newtons laws of motion is that:\n\n"for an object that is stationary or travelling at constant velocity, no force acts."\n\nso you ask what is the force?\nthere is no resultant force, the driving force and resistance are equal and cancel each other out.\n\n\nAnd additionally the basic equation to find force is:\nF=ma (force = mass x accelleration), this is evidence that an accelleration needs to be taking place for a force to act.\n\nLets say you want the force for a 50kg object accellerating at 3ms-2. F=ma  ->  F=50x3 ->  F=150N.\n\nNow lets say you want the force for a 50kg object travelling at constant velocity, no accelleration.\nF=ma  ->  F=50x0  ->  F=0N\n\n\nThere you go, wasnt that easy
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.
718152
weight equals mass times acceleration due to gravity
831091
w = mg\n(w is weight, m is mass, g is acceleration due to gravity (9.8m/s^2))\n :. m = w/g\n        = 300 / 9.8\n        = 30.6 kg
834482
The answer is C. Although B seems to be a possible answer at first, there could still be an n-number of forces acting on the object such that the sum of all the force vectors is zero, thus making C the more complete answer. One of Newton's laws says that F=m*a (force on an object equals its mass times its acceleration), or a force of 1 Newton on a 1kg object will make it accelerate at a rate of 1m/s^2 in the same direction of the force.