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Lelu [443]
3 years ago
5

A spherical weather balloon is filled with hydrogen until its radius is 4.40 m. Its total mass including the instruments it carr

ies is 19.0 kg. (a) Find the buoyant force acting on the balloon, assuming the density of air is 1.29 kg/m3. kN (b) What is the net force acting on the balloon and its instruments after the balloon is released from the ground? kN (c) Why does the radius of the balloon tend to increase as it rises to higher altitude?
Physics
1 answer:
ipn [44]3 years ago
5 0

Answer:

4515.49484 N

4329.10484 N

Explanation:

r = Radius of balloon = 4.4 m

m = Mass of balloon with instruments = 19 kg

g = Acceleration due to gravity = 9.81 m/s²

Volume of balloon

v=\frac{4}{3}\pi r^3\\\Rightarrow v=\frac{4}{3}\pi 4.4^3\\\Rightarrow v=356.8179\ m^3

The Buoyant force = Weight of the air displaced

F=\rho vg\\\Rightarrow F=1.29\times 356.8179\times 9.81\\\Rightarrow F=4515.49484\ N

The buoyant force acting on the balloon is 4515.49484 N

Net force on the balloon

F_n=F-W\\\Rightarrow F_n=4515.49484-19\times 9.81\\\Rightarrow F_n=4329.10484\ N

The net force on the balloon is given by 4329.10484 N

As the balloon goes up the pressure outside reduces as the density of air decreases while the air pressure inside the balloon is high hence, the radius of the balloon tend to increase as it rises to higher altitude.

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Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

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Using law of conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

Along X- axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}

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Along Y-axis

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Negative sign shows the direction of first ball .

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