Physics, asked by rkarni680, 2 months ago

define terminal velocity and explain expression​

Answers

Answered by shubhamgariya660
1

Answer:

Terminal velocity is defined as the highest velocity attained by an object that is falling through a fluid. It is observed when the sum of drag force and buoyancy is equal to the downward gravity force that is acting on the object. The acceleration of the object is zero as the net force acting on the object is zero.

Answered by Anonymous
1

Answer:

\huge\underline\mathfrak\red{❥︎ANSWER}

Terminal velocity is defined as the highest velocity attained by an object that is falling through a fluid. It is observed when the sum of drag force and buoyancy is equal to the downward gravity force that is acting on the object. The acceleration of the object is zero as the net force acting on the object is zero.

Where,

b: constant depending on the type of drag

∑F=ma (free fall of an object)

mg−bv2=ma (assuming that the free fall is happening in positive direction)

mg−bv2=mdvdt 1mdt=dvmg−bv2 (differential form of the equations)

∫1mdt=∫dvmg−bv2 (integrating the equations)

∫dvmg−bv2=1b∫dvα2−v2

Where,

α=mgb−−−√ dv=αsech2(Θ)dΘ (after substituting for v=αtanh(Θ))

v2=α2tanh2(Θ)

After integration

1b∫αsech2(Θ)dΘα2−α2tanh2(Θ) 1b∫αsech2(Θ)dΘα2(1−tan2Θ) 1b∫αsech2(Θ)dΘα2sech2(Θ)=1αb∫dΘ=1αbarctanh(vα)+C (using the identity 1−tanh2(Θ)=sech2(Θ)

Where,

b: constant depending on the type of drag

∑F=ma (free fall of an object)

mg−bv2=ma (assuming that the free fall is happening in positive direction)

mg−bv2=mdvdt 1mdt=dvmg−bv2 (differential form of the equations)

∫1mdt=∫dvmg−bv2 (integrating the equations)

∫dvmg−bv2=1b∫dvα2−v2

Where,

α=mgb−−−√ dv=αsech2(Θ)dΘ (after substituting for v=αtanh(Θ))

v2=α2tanh2(Θ)

After integration

1b∫αsech2(Θ)dΘα2−α2tanh2(Θ) 1b∫αsech2(Θ)dΘα2(1−tan2Θ) 1b∫αsech2(Θ)dΘα2sech2(Θ)=1αb∫dΘ=1αbarctanh(vα)+C (using the identity 1−tanh2(Θ)=sech2(Θ))

hope it helps you ☺☺

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