The drag force F on a boat varies jointly with the wet surface area A of the boat and the square of the speed s of the boat. A b
1 answer:
You might be interested in
Answer:
The fraction of mass that was thrown out is calculated by the following Formula:
M - m = (3a/2)/(g²- (a²/2) - (ag/2))
Explanation:
We know that Force on a moving object is equal to the product of its mass and acceleration given as:
F = ma
And there is gravitational force always acting on an object in the downward direction which is equal to g = 9.8 ms⁻²
Here as a convention we will use positive sign with acceleration to represent downward acceleration and negative sign with acceleration represent upward acceleration.
Case 1:
Hot balloon of mass = M
acceleration = a
Upward force due to hot air = F = constant
Gravitational force downwards = Mg
Net force on balloon is given as:
Ma = Gravitational force - Upward Force
Ma = Mg - F (balloon is moving downwards so Mg > F)
F = Mg - Ma
F = M (g-a)
M = F/(g-a)
Case 2:
After the ballast has thrown out,the new mass is m. The new acceleration is -a/2 in the upward direction:
Net Force is given as:
-m(a/2) = mg - F (Balloon is moving upwards so F > mg)
F = mg + m(a/2)
F = m(g + (a/2))
m = F/(g + (a/2))
Calculating the fraction of the initial mass dropped: