Answer:
To find the quickest path to the ground
Explanation:
Answer:
Δt=0.85 seconds
Explanation:
In this chase the speed does not change as the mass change.So we can use the follow equation to find the required time
Δt=Δv/gμ
To stop the final speed will be zero therefore the change in speed will be
Δv= vf-vi
Δv=0-5 m/s
Δv= -5 m/s
Now we plug our values for Δv,g and μ to find time
Δt=Δv/gμ
Δt=(-5m/s) ÷(9.8m/s² × 0.6)
Δt=0.85 seconds
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
Answer:
time of flight of a pulse, and these most often
involve triggering of the measuring oscilloscope
with the signal that generates the sound pulse and
timing the time delay of the pulse picked up by a
conveniently placed microphone45
. Loren Winters
has reported a method similar in principle to the
present one, but which uses a completely different
detection system6
.
Explanation:
