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The viscosity of air provides a drag force on a raindrop and keeps it from falling with the acceleration of gravity. When a drop is falling (assuming it does not combine with other drops in the process) it will reach a terminal velocity which depends on its diameter. The larger the diameter the larger the terminal velocity. Specifically, the terminal velocity is proportional to the square root of the diameter of the drop. Big rain drops fall faster than small rain drops. See related links for details and equations.
If the raindrop is falling at a constant speed, then it has reached terminal velocity. This happens when the downward force (due to gravity) is the same as the upward force due to friction. As such the net force acting on the rain drop is 0.
Considering the rain droplets as spherical body. We have two forces acting on the rain drop when it is falling through the sky, namely the resistance force due to friction(drag force)upwards and its weight downwards. Th rain drop falling from such distance attain a terminal velocity while falling i.e their speed becomes constant after sometime. This happens when the drag force equals the weight of drop,, this happens because drag force increases with velocity of the drop. Drag force= .5*rho*v2*A(frontal area)*Cd(coefficient of drag) Weight=m*g=rho*volume of spherical drop=rho*4/3*r3. When we equalize it, we get the Terminal Velocity(v) varying directly as sqr of r(radius of drop) So larger drop means, larger terminal velocity and hence less time taken for falling. So larger rain drop falls faster.
It does, up to a limit called "terminal velocity". Terminal velocity is reached when the force of friction against the air equals the force of gravity acting on the raindrop. As the drop falls, it hits molecules in the air, and each of those molecules slows the raindrop down just a little bit. As gravity pulls the drop down, soon it hits so many molecules per unit of time that the combined effect prevents it from gaining any more speed.
the reason why is because of the shape and weight if the rain drop. the weight of a rain drop is less than a gram and the shape of one when falling from the sky, has a lot of drag. this means that the terminal velocity of a rain drop is very slow
A falling body initially falls at a rate of -9.8m/s2, the acceleration due to gravity. Because of the drag force of the air, which is an upward force that opposes the force of gravity, the body's acceleration will decrease as it continues falling. When the drag force equals the weight of the falling body, there will be no further acceleration, and the body will have reached terminal velocity.
Yes, if you were to drop a rock, after one second has elapsed, gravity would make it move from a standstill to its terminal velocity in the medium through which it is falling. In general, the speed would change at a rate of 9.8 m/s2.
The viscosity of air provides a drag force on a raindrop and keeps it from falling with the acceleration of gravity. When a drop is falling (assuming it does not combine with other drops in the process) it will reach a terminal velocity which depends on its diameter. The larger the diameter the larger the terminal velocity. Specifically, the terminal velocity is proportional to the square root of the diameter of the drop. Big rain drops fall faster than small rain drops. See related links for details and equations.
If the raindrop is falling at a constant speed, then it has reached terminal velocity. This happens when the downward force (due to gravity) is the same as the upward force due to friction. As such the net force acting on the rain drop is 0.
The velocity of free falling bodies do change However there are some exceptions like a free falling rain drop Please mention the case of which you want to know too
Absolutely,Although the effect will be minimal if you drop the quarter from waist height.If you drop it from an airplane, it might even reach terminal velocity where the air resistance would counteract and balance the acceleration due to gravity.
Due to gravity, the bullet starts to drop the second it leaves the barrel. You can calculate the drop by factoring mass and velocity with gravity (9.8 m/s²).
Considering the rain droplets as spherical body. We have two forces acting on the rain drop when it is falling through the sky, namely the resistance force due to friction(drag force)upwards and its weight downwards. Th rain drop falling from such distance attain a terminal velocity while falling i.e their speed becomes constant after sometime. This happens when the drag force equals the weight of drop,, this happens because drag force increases with velocity of the drop. Drag force= .5*rho*v2*A(frontal area)*Cd(coefficient of drag) Weight=m*g=rho*volume of spherical drop=rho*4/3*r3. When we equalize it, we get the Terminal Velocity(v) varying directly as sqr of r(radius of drop) So larger drop means, larger terminal velocity and hence less time taken for falling. So larger rain drop falls faster.
It does, up to a limit called "terminal velocity". Terminal velocity is reached when the force of friction against the air equals the force of gravity acting on the raindrop. As the drop falls, it hits molecules in the air, and each of those molecules slows the raindrop down just a little bit. As gravity pulls the drop down, soon it hits so many molecules per unit of time that the combined effect prevents it from gaining any more speed.
Unless you drop the feather in a vacuum, air resistance will be significant, so any acceleration (change in velocity) will not be due solely to gravity.
When given a constant acceleration, just multiply it by time, t, to detemine the final velocity. If the initial velocity was zero (as is the case when you drop something), then the average velocity is half the terminal velocity.
The determination of charge of an electron was done by R.A. Millikan in 1913, In R.A. Millikan's working (Millikan's oil drop experiment) explained as follows the terminal velocity of charged oil drop is measured under two conditions viz. under gravity alone and under the combined action of gravity and electric field oppose the gravity . Negatively charge drop is selected and its motion is studied.I think it's due to this electron has (-)ve charge.In fact, it is a very long experiment ,i can't explain all of it.