a hot air balloon weighing 30N is tied to the ground by a string to prevent from floating off the ground. The volume of the ball
oon is 20m³ and the density of air is 1.3kgm-³. find fhe force exerted by the rope on the balloon
1 answer:
Answer:
The force exerted by the rope is FT = 225.06 [N]
Explanation:
In order to solve this problem we must use a static analysis, since Globe does not move. For a better understanding in solving this problem, a free body diagram with the forces acting on the globe is attached.
The buoyant force acts upward as it causes the balloon to tend to float, the weight of the balloon tends to lower the balloon and the downward tension force does not allow the balloon to float
The buoyant force is defined by the following equation:
FB = Ro*V*g
where:
FB = Buoyant force [N]
Ro = density of the air = 1.3 [kg/m^3]
V = volume of the balloon = 20 [m^3]
g = gravity acceleration = 9.81 [m/s^2]
FB = 1.3*20*9.81 = 255.06 [N]
Now we do a sum of forces equal to zero in the y-axis
FB - 30 - FT = 0
255.06 -30 = FT
FT = 225.06 [N]
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Answer:
The magnitude of the large object's momentum change is 3 kilogram-meters per second.
Explanation:
Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:
(1)
Where:
,
- Initial and final momemtums of the small object, measured in kilogram-meters per second.
,
- Initial and final momentums of the big object, measured in kilogram-meters per second.
If we know that ,
and
, then the final momentum of the big object is:
The magnitude of the large object's momentum change is:
The magnitude of the large object's momentum change is 3 kilogram-meters per second.