A top of rotational inertia 4.0 kg m2 receives a torque of 2.4 nm from a physics professor. the angular acceleration of the body
1 answer:
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Answer:
The drag coefficient is
Explanation:
From the question we are told that
The density of air is
The diameter of bottom part is
The power trend-line equation is mathematically represented as
let assume that the velocity is 20 m/s
Then
The drag coefficient is mathematically represented as
Where
is the drag force
is the density of the fluid
is the flow velocity
A is the area which mathematically evaluated as
substituting values
Then
I attached a picture of the diagram associated with this question.
Now,
When we check the vertical components of the tension in the rope, we will find that we have two equal components acting upwards.
These two components support the weight and each of them has a value of TcosΘ
The net force acting on the body is zero.
Fnet=Force of tension acting upwards-Force due to weight acting downwards
0 = 2TcosΘ -W
W = 2TcosΘ
T = W / 2cosΘ
Now,
When we check the vertical components of the tension in the rope, we will find that we have two equal components acting upwards.
These two components support the weight and each of them has a value of TcosΘ
The net force acting on the body is zero.
Fnet=Force of tension acting upwards-Force due to weight acting downwards
0 = 2TcosΘ -W
W = 2TcosΘ
T = W / 2cosΘ