When a falling object reaches terminal velocity, the net force acting on it is ____?
152239
AIR RESISTANCE CANCELS OUT THE FORCE DUE TO GRAVITY , THERE IS NO ACCELERATION BECAUSE THE OPPOISTE FORCE IN THE SAME.\nSO YOU CAN SAY 0
942588
zero, \n\nno acceraltion = no net force\n\nmathematically  Net F = ma\n\nif the right side is 0, left side is too
981397
This might be one of those questions that you have to answer with just using the formula, i.e. you will not get an actual answer, but an equation that if you knew the mass, length and force you could calculate the height dependant upon the force input, but thats pretty high level stuff, you would need so solve for angular acceleration and momentum, also in this sort of question you would be given either an impact force on a part of the rod (i.e. where the rod was hit) of you would be told the rod has some momentum. Then you could solve.\n\nI suggest you have another look at the question, I think you will find it says the rod is already moving.
730176
All of these questions can be answered using Newton's 2nd Law:\n\n"The rate of change of momentum of a body is proportional to the resultant force acting on the body and is in the same direction."\n\nThus, if a box has a net force of 0 N, it will not accelerate. This means that its velocity will not change. Note that this DOES NOT necessarily mean that its velocity is 0, just that the velocity is not changing.\n\nIf the box has a net force of 15 N, then it will accelerate in the direction of the applied force. Note that the magnitude of the acceleration is dependant upon the mass of the box.\n\nSo, to answer your two questions:\n\nA) Yes, it will have constant velocity.\n\nB) Yes, the box will accelerate.
30290
draw the free body diagram. ok, there's a force of 21N going down and an unknown force going up. you know that the net force is 12N. so that means how much of the 21N force is being cancelled out by the buoyant force to produce the new weight? so it's just 21N-12N which is 9N
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).
217280
According to Newton's second law,\nF = m*a \nF = Force\nm = mass\na = acceleration\n\nnow, also,\n\na = (v-u)/t \nv = final velocity\nu = initial velocity\nt = time\n\nput this expression into first equation, you get\n\nF = m (v - u)/t\n= (mv - mu)/t\n\nalso, momentum = mass * velocity\n\nso,\nF = (final momentum - initial momentum)/t\n\nwe can also say that\n\nF = Rate of change of momentum
831377
it is the force of gravity it's simply 9.8 m/s/s or 32ft/s/s     \nacceleration of gravity is -9.8m/s/s or 32ft/s/s\ng force is just the absolute value
774794
Remember the net work is the change in Kinetic energy.  In order for a force to do work, it must be parallel the motion.\n\nW = f (dot) d   =  f*d*cos(theta)\n\n Also, Mechanical (and Total) energy is conserved. Thus, any increase in kinetic energy is coupled with an equal decrease in potential energy. \n\n/&#92;K = -/&#92;U = Wnet
716944
If there is no net force (i.e. sum of forces is zero) then there will be no change in velocity. So if the object was going 15 m/s and the net force is zero, then it will still be going 15 m/, (for 2 sec = 30 meters traveled). Same concept for your jet car - constant velocity means no net force ( there may be forces acting on it, but if the velocity is constant, they must sum to zero - net force is zero).\n\nAs Newton's first law says, "An object in motion tends to stay in motion unless acted on by an outside force." So constant velocity (i.e. stays in motion at the same velocity) means no net force acting on the object.
694476
In your lab you are dealing with the relationship of:\nf=m*a\nso a=f/m\n\nIf you plot the values of acceleration versus mass holding force constant, this is a non-linear graph, so it will depend on the range of your data.  As m gets large, the slope will approach zero.  As m gets small the slope will approach infinity, which makes sense.  The instantaneous slope of this line is -f/m^2.  So your error will depend on the relative magnitude of the mass with respect to the net force.\nIf you plot the values of acceleration versus net force while holding the mass constant, the slope of this line is 1/m.  Your error will depend on the accuracy of the system.  If the coefficient of kinetic friction is not a constant, or the force applied is not a constant, then error will be introduced into the system.\n\nj
250025
The net force is zero.  Remember the equation F=MA:  force equals mass times acceleration.  If something is at rest, A is zero, so F must be zero.\n\nThe gravitational force on the brick is exactly balanced by the force of the table holding the brick up.  Net force is zero.
818860
I'm not going to give you the complete answer because you'll learn more by working through it yourself, but I'll give a few pointers.\n\nAny object travelling in a circle at a constant speed has a centripital acceleration that can be calculated from the velocity and the radius of the circle.\n\nNewton tells us that F[net] = ma, so the object must be experiencing a net centripetal force equal to its mass multiplied by its centripetal acceleration.\n\nWhere does this force come from?  In your problem, it sounds as though the only force that could be pulling the box toward the center of the circle (the centripetal force) is the friction between the box and the bed of the truck.  So, when you add up all the (horizontal) forces to get the net force, the olny one is going to be friction.  F[net] = F[friction] \n\nNow all you need to do set F[friction]  equal to the F[net] that you calculated above (remember that F[net] was equal to your centripetal force), and solve for the coefficient of friction (your textbook shold have the formula for static friction, which will include the coefficient of static friction).\n\nGood luck!
252215
2nd Law of motion:  the net force = the time derivative of the momentum vector.
900616
If the rock is not being slowly pushed into the ground , then the net force is 0 because the force down is equal to the normal force upwards exerted by the ground.\n\nif the car is really heavy and pushes the rock into the ground, then the net force is greater than 0, in a downwards direction.
468173
it depends on how much the ground "gives". when it falls from a known height, it has potential enrgy=mgh, which translates to kinetic energy 1/2mv^2, so impact velocity is v = sqrt(2gh). final velocity (assuming inelastic collision) is 0. if you know how long it takes to deaccelerte to 0, the average force is f=ma=m*v/t = m*sqrt(2gh)/t. if you know how far it travels before stopping, v^2 = 2ax, a=v^2/2x, f=ma=mgh/x\n\n\ni realize an easier way to do this is to say that the work done by the ground is the same as the force on the object times the distance traveled through the ground. since \nWork = change in Energy\nEnergy = mgh\ndistance into ground = x,\nW = Fx\nW= E = mgh\nmgh= Fx\nF = mgh/x
613926
Why do people cite "Myth busters" as if it is some type of scientific source?\nIt is a TV show...and not a very good one at that.  Sometimes they get it "right", sometimes they are wrong.\nJustifying something because "they did it on Myth busters" is not a good idea if you want to be taken seriously.\n\nThe principle at work here is called terminal velocity.\nIn a vacuum, when an object is dropped, it continues to accelerate as it falls, its velocity ever increasing until it hits the ground.  However, most things (including the Empire State Building) are not in a vacuum, and thus, air resistance comes into play.  After the object is dropped, it will accelerate due to gravity.  As the object accelerates and its velocity increases, the frictional force acting on the object increases.  There will come a point when the frictional (upward) force exerted on the object is balanced by the objects weight (downward force).  When the forces are balanced, no acceleration takes place and the object maintains a constant velocity....called the terminal velocity.\n\nThe terminal velocity depends on several factors, two of which being the cross-sectional surface area the air pushes against, another being the mass of the object.\nA higher cross sectional area allows a higher frictional force.  A lower mass means the force applied will have a greater effect on accelerating the object due to Newton's 2nd law (F = ma).\nIn the case of a penny, the terminal velocity is probably pretty low because it has such a small mass with a relatively high surface area.  It would not take long for the penny to reach its terminal velocity and stop accelerating.  At this point, the penny is fallings as fast as it every will naturally through the air due to gravity.