Many birds can attain very high speeds when diving. Using radar, scientists measured the altitude of a barn swallow in a vertica
l dive; it dropped 208 m in 3.0 s. The mass of the swallow was estimated to be 0.018 kg, and its cross-section area as 5.6×10−4m2. What was the drag coefficient for this swallow as it dove?
1 answer:
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Answer:
The torque on the child is now the same, τ.
Explanation:
- It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
- In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
- The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.
- When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:
- So, τ = 3/2*m*r²*α (2)
- When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:
- Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:
τ = 3/4*m*r² * (2α) = 3/2*m*r²
same result than in (2), so the torque remains the same.
Answer:
Explanation:
Given that:
Absolute temperature of the body,
- emissivity of the body,
<u>Using Stefan Boltzmann Law of thermal radiation:</u>
where:
(Stefan Boltzmann constant)
Now putting the respective values: