46.5k views
4 votes
a balloon has 2 g of air. a small hole is pierced into it. the air comes out with a velocity of 4 m/s. if the balloon shrinks conpletely in 2.5s, theaverage force acting on the balloon is?solution

User Timat
by
7.4k points

1 Answer

4 votes

Final answer:

The average force acting on the balloon as air escapes through a small hole is calculated using Newton's second law and the information provided. The force is found to be 0.0032 Newtons.

Step-by-step explanation:

The question asks to determine the average force acting on a balloon as the air inside escapes through a small hole. Given that the balloon has 2 g of air (mass), and it deflates completely in 2.5 seconds, with the air escaping at a velocity of 4 m/s, we can calculate the force using Newton's second law. To solve this problem, we'll need to use the formula F = Δp/Δt (where p is momentum and t is time), and the fact that the momentum change (Δp) is equal to the mass of the air multiplied by the velocity (m*v).

First, we calculate the momentum change of the air: Δp = m*v = 0.002 kg * 4 m/s = 0.008 kg*m/s.

Next, we calculate the average force exerted on the balloon: F = Δp/Δt = 0.008 kg*m/s / 2.5 s = 0.0032 N.

Therefore, the average force acting on the balloon is 0.0032 Newtons.

User Porter
by
9.1k points