Richard "Ditch" Brodie
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And terminal velocity is computed in meter per second i.e \(ms^{-1}\) Solved Examples for Terminal Velocity Formula. Q.1: A man is at the height of 2000 m from the ground. What would be his terminal velocity using Terminal Velocity Formula? Solution: Given, Height h = 2000 m, The terminal velocity formula is: \(V_{Terminal} = \sqrt{2 g h}\)
The distance travelled by the ball in t = 0.2s after it attains terminal velocity is View Solution Click here:point_up_2:to get an answer to your question :writing_hand:derive expression for terminal velocity when a ball of radius r is dropped thrugh a
The terminal velocity of an object is the maximum constant velocity acquired by the object while falling ...
A solid sphere falls with a terminal velocity of 20 m s −1 in air. If it is allowed to fall in vacuum, (a) terminal velocity will be 20 m s −1 (b) terminal velocity will be less than 20 m s −1 (c) terminal velocity will be more than 20 m s −1 (d) there will be no terminal velocity
The terminal velocity of a liquid drop of radius r falling through air is v. If two such drops are combined to form a bigger drop, the terminal velocity with which the bigger drop falls through air is (Ignore any buoyant force due to air) : √ 2 v; 2 v; 3 √ 4 v; 3 √ 2 v
This is the formula for terminal velocity. It is apparent from the above expression that the terminal velocity is (a) directly proportional to the square of the radius of body, (b) directly proportional to the densities of the body and the medium, (conversely proportional to the coefficient of viscosity of the medium.
After some time the velocity of the ball attains a constant value known as terminal velocity v T. The terminal velocity depends on ( i ) the mass of the ball m , ( i i ) η , ( i i i ) r , and ( i v ) acceleration due to gravity g .
Terminal velocity is the maximum velocity attainable by an object as it falls through a fluid. It occurs when the sum of the drag force and the buoyancy is equal to the downward force of gravity acting on the object.
The terminal velocity of a steel ball 2 mm in diameter falling through glycerin is 44 × 10 − 2 c m / s (Given that specific gravity of steel = 8, specific gravity of steel =8, specific gravity of glycerin a 1.3, viscosity of glycerine 8.3 poise.)
Eight drops of water, each of radius 2 mm are falling through air at a terminal velocity of 8 c m s − 1. If they coalesce to form a single drop, then the terminal velocity of combined drop will be: View Solution
