Flapping flight is very energy intensive. A wind tunnel test
1 answer:
Answer:
The metabolic power for starting flight=134.8W/kg
Explanation:
We are given that
Mass of starling, m=89 g=89/1000=0.089 kg
1 kg=1000 g
Power, P=12 W
Speed, v=11 m/s
We have to find the metabolic power for starting flight.
We know that
Metabolic power for starting flight=
Using the formula
Metabolic power for starting flight=
Metabolic power for starting flight=134.8W/kg
Hence, the metabolic power for starting flight=134.8W/kg
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Refer to the diagram shown below.
u = 0, the initial vertical velocity
Assume g = 9.8 m/s² and ignore air resistance.
At the first stage of landing on the ground, the distance traveled is
h = 3.1 - 0.6 = 2.5 m.
If v = the vertical velocity at this stage, then
v² = u² + 2gh
v² = 2*(9.8 m/s²)*(2.5 m) = 49 (m/s)²
v = 7 m/s
At the second stage of landing on the ground, let a = the acceleration (actually deceleration) that his body provides to come to rest.
The distance traveled is 0.6 m.
Therefore
0 = (7 m/s)² + 2(a m/s²)*(0.6 m)
a = - 49/1.2 = - 40.833 m/s²
Answers:
(a) The velocity when the man first touches the ground is 7.0 m/s.
(b) The acceleration is -40.83 m/s² (deceleration of 40.83 m/s²) to come to rest within 0.6 m.
u = 0, the initial vertical velocity
Assume g = 9.8 m/s² and ignore air resistance.
At the first stage of landing on the ground, the distance traveled is
h = 3.1 - 0.6 = 2.5 m.
If v = the vertical velocity at this stage, then
v² = u² + 2gh
v² = 2*(9.8 m/s²)*(2.5 m) = 49 (m/s)²
v = 7 m/s
At the second stage of landing on the ground, let a = the acceleration (actually deceleration) that his body provides to come to rest.
The distance traveled is 0.6 m.
Therefore
0 = (7 m/s)² + 2(a m/s²)*(0.6 m)
a = - 49/1.2 = - 40.833 m/s²
Answers:
(a) The velocity when the man first touches the ground is 7.0 m/s.
(b) The acceleration is -40.83 m/s² (deceleration of 40.83 m/s²) to come to rest within 0.6 m.