Is the total acceleration betwen the earth and the falling body dependent on bodies mass?
408862
the force between the object and the earth is dependant on mass, but the acceleration of the object towards the earth is 9.81 meters&#92;second. (ignoring air resistance)\n\nforce = mass x acceleration
60690
The amount of matter in an object is called its mass. The force of gravity is dependent on the amount of mass a body has. That means that the gravity on the Earth is greater than the gravity on the Moon, since the Earth has much more matter or mass than the Moon.
611971
Mass is a constant; weight varies by gravitational pull.
912569
I think F is the answer. Taking acceleration due to gravity is constant at 9.81ms^-2, but this assumes that air resistance is neglected, and g is a constant.\n\nIf air resistance is NOT neglected, and acceleration due to gravity is not taken to be constant, H is the correct answer. The rest of the options are wrong.
324067
No. Gravity has to do with the mass of objects--you have some too.
247359
Mass interacting with gravity
913200
Both their accelerations will be the same and equal to the acceleration due to gravity, g=9.81 m/s/s...This is coz the only force acting on them is the force due to gravity
135033
Probably won't work.  If you assume 10 billion people with an average weight of 150 lbs, the mass of the earth is about 10,000,000,000,000 (10^13) times greater.  If everyone jumped up and down, even on the same side of the Earth, it wouldn't affect the earth's momentum or motion in it's orbit.  The acceleration applied by the mass of people is just too small.  Think of it as being in the 13th decimal place(0.0000000000001), it's essentially unnoticeable.
514076
It's true - gravity pulls harder (i.e. more force) on a more massive (i.e. heavier) object. BUT that object is harder to accelerate - it has more interia. The two effects cancel each other out.\n\nClearly you don't like equations, but putting it in numbers can help:\n\nIf a 1kg object is pulled towards the earth, what will be the force, and what will be the acceleration?\nForce = mass * earth's gravitational field strength\nso Force = 1kg * 9.81N/kg\nForce = 9.81N (or 10N, roughly)\n\nAcceleration = Force / Mass\nSo Acc'n = 9.81N / 1kg = 9.81 m/s^2\n\nNow consider the same for a 10kg object:\n\nForce is now 98.1N (or 100N, roughly)\nSo Acc'n = 98.1N / 10kg = 9.81 m/s^2\n\nSo the acceleration is the same.
127639
the acceleration of an object is dependent upon both force and mass...Forces cause acceleration\n\nit takes an action force to cause acceleration for the object\n\naccelerate if there is an unbalanced force acting upon it.
896953
The net force is the difference between the two weights.\n\nHowever, there is not enough information in the question to answer without making some assumptions.\n\nYou are not given the mass of the man and/or the bricks. One can calculate the mass if given the force of gravity, however you aren't given that either.\n\nOne method of solving is:\n\nASSUMING that this system is on the surface of the Earth, you can find the two masses, using F=mg with g=9.8N/kg. Once you've got that you can find the acceleration using F=ma since you know the net force and the mass of the system.\n\nMake sure you indicate which direction the acceleration is in (in this case, which object is falling, and which is being raised.
172859
The terminal velocity of a falling body occurs during free fall when a falling body experiences zero acceleration. This is because of the retarding force known as air resistance. Air resistance exists because air molecules collide into a falling body creating an upward force opposite gravity. This upward force will eventually balance the falling body's weight. It will continue to fall at constant velocity known as the terminal velocity.\n\nThe magnitude of terminal velocity depends on the weight of the falling body. For a heavy object, the terminal velocity is generally greater than a light object. This is because air resistance is proportional to the falling body's velocity squared. For an object to experience terminal velocity, air resistance must balance weight. An example that shows this phenomenon was the classic illustration of a rock and a feather being dropped simultaneously. In a vacuum with zero air resistance, these two objects will experience the same acceleration. But on the earth this is not true. Air resistance will equal weight more quickly for the feather than it would for the rock. Thus the rock would accelerate longer and experience a terminal velocity greater than the feather.\n\nAnother factor that affects terminal velocity is the orientation at which a body falls. If an object falls with a larger surface area perpendicular to the direction of motion it will experience a greater force and a smaller terminal velocity. On the other hand, if the object fell with a smaller surface area perpendicular to the direction of motion, it will experience a smaller force and a greater terminal velocity.\n\nThe terminal velocity for a skydiver was found to be in a range from 53 m/s to 76 m/s. Four out of five sources stated a value between 53 m/s and 56 m/s. Principles of Physics stated a value of 76 m/s. This value differed significantly from the others. Then again, the value is variable since the weight and the orientation of the falling body play significant roles in determining terminal velocity.\n\nAn object which is falling through the atmosphere is subjected to two external forces. One force is the gravitational force, expressed as the weight of the object. The other force is the air resistance, or drag of the object. The motion of any object can be described by Newton's second law of motion, force F equals mass m times acceleration a: \n\nF = m * a \n\nwhich can be solved for the acceleration of the object in terms of the net external force and the mass of the object: \n\na = F / m \n\nWeight and drag are forces which are vector quantities. The net external force F is then equal to the difference of the weight W and the drag D \n\nF = W - D \n\nThe acceleration of a falling object then becomes: \n\na = (W - D) / m \n\nThe drag force depends on the square of the velocity. So as the body accelerates its velocity and the drag increase. It quickly reaches a point where the drag is exactly equal to the weight. When drag is equal to weight, there is no net external force on the object and the object falls at a constant velocity as described by Newton's first law of motion. The constant velocity is called the terminal velocity .\n\nWe can determine the value of the terminal velocity by doing a little algebra and using the drag equation. Drag depends on a drag coefficient, Cd the air density, r the square of the velocity V and some reference area A of the object: \n\nD = Cd * r * V ^2 * A / 2 \n\nAt terminal velocity, D = W. Solving for the velocity, we obtain the equation \n\nV = sqrt ( (2 * W) / (Cd * r * A) ) \n\nwhere sqrt denotes the square root function. Typical values of the drag coefficient are given on a separate slide.\n\nThe terminal velocity equation tells us that an object with a large cross-sectional area or a high drag coefficient falls slower than an object with a small area or low drag coefficient. A large flat plate falls slower than a small ball with the same weight. If we have two obje
39375
The force of gravity is actually the weakest of the Fundamental Forces.  But to answer your question,  our heads don't go crashing to the floor because our bodies are sufficiently strong enough to apply enough 'upward' preassure to counter-act the Earth's gravitational attraction.  Every time you jump into the air, you're essentially escaping Earth's gravity, though it's just for a moment.  So, you can see, it doesn't take much effort for something as small as a body (or a head) to overcome gravity.
870312
a)For rigid body, center of mass accelerates in the direction of applied external forces. Just Newton's second law and therefore true.\nb)As in Newton's second law F=ma, force is dependent on acceleration and not directly to velocity only. So, there is explicit time dependence. hence, it really matters how fast the object gets to 10m/s or 20m/s velocity starting from rest. Therefore this statement is inconclusive with unknow time.
738834
d is the most correct answer, yes.  \n\nHowever keep in mind that despite g being the "gravitational constant", acceleration caused by gravity DOES change depending on the the distance from the earth.  When we're talking about the difference of a table, there is practically no difference, however keep in mind that it does change.
22095
Let's try to anser the question without looking it up and without using a calculator.\n\nWe know Newton's gravitation law: F = G*m*M/r^2\nAnd we know the gravity on earth: F = m*g\n==> m*g = G*m*M/r^2\n==> M = r^2*g/G\nNow, the acceleration on the earth surface is approximately 10 m/s^2\nThe earth's radius is approximately 6*10^6 m\nAnd I remember G to be approximately 7*10^-11 Nm^2/kg^2\n\nM = 6^2*10/7 * 10^12/10^-11 kg\n\nThe earth weights about 5*10^24 kg\n\nIf you look it up in Wikipedia, it says 5,9736*10^24 kg. \n\nSo we're close enough if you take into account that we used rough estimates for the constants.
69858
Weight is a force.  The National Institute of Science and Technologiy defines it here as such.\nhttp://physics.nist.gov/Pubs/SP811/sec08.html\n\nA force is acceleration times mass.  Weight is the accleration due to gravity times mass.  On the Earth's surface, this is approximately:\n\nWeight = Mass x 9.8 meters/second^2\n\nFor a accurate conversion, you would need to use the acceleration due gravity minus the accelration due to centrifugal force at your exact location.  \n\nThis is why physicists more commonly use mass than weight to describe an object.