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# the force of buoyancy exerted by the atmosphere on a balloon is B in upward direction and remains constant . the force of air resistance on the balloon acts opposite tom the direction of velocity and is proportional to it .The balloon carries a mass M and is found to fall down near the earths surface with a constant velocity v . How much mass should be removed from the balloon so that it may rise with a constant velocity  v ?

Asked by geethamehs08 19th October 2022, 6:28 PM
When the balloon is descending with constant speed v , net force acting on balloon is zero
by Newton's second law of motion .

Net Force :-  B + ( k v ) - ( M g ) = 0    ..........................(1)

where B is buoyant force , ( kv) is drag force that is directly proportional to velocity v ,
M is initial quantity of mass carried by the balloon and g is acceleration due to gravity

When the balloon is ascending with constant speed v , net force acting on balloon is zero by
Newton's second law of motion .

Net Force :-  B  -  ( k v ) - ( m g ) = 0    .........................(2)

where  m is the new quantity of mass carried by the balloon and g is acceleration due to gravity.

( drag force 'kv' is opposite to the direction of motion )

By subtracting eqn.(2) from eqn.(1) , we get

2 ( k v ) = ( M - m ) g

(M - m ) =  [ 2 (k v ) ] / g

Hence , amount of mass to be removed to make the balloon ascending is (M - m ) =  [ 2 (k v ) ] / g
Answered by Expert 20th October 2022, 4:42 PM
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