The terminal velocity as it falls through still air is 4.65154 in/s.
The diameter of small water droplet is 1.25 mil= 1.25×0.0254×10^-3 m
= 3.175 × 10^-5 m
Now the viscosity of still air is η = 1.83× 10⁻⁵ Pa
So the formula for drag force is:
Fd = 6πηrv
where, v is the velocity.
Now to attain terminal velocity acceleration must be zero.
→ W = Fd
ρVg = 6πrηv
ρ × 4/3 πr³×g = 6πrηv
v = 2/9 × ρgr³/ η
v = 2/9 × 10³×9.81×(3.175×10^-3) / 18.6×10^-6
v = 0.1181 m/s
v = 4.65154 in/s
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Answer:
7 m/s^2
Explanation:
Given that the jet is traveling 37.6 m/s when the pilot receives the message.
And it takes the pilot 5.37 s to bring the plane to a halt.
Acceleration of the plane can be calculated by using first equation of motion
V = U - at
Since the plane is going to stop, the final velocity V = zero.
And the acceleration will be negative
Substitute all the parameters into the formula
0 = 37.6 - 5.37a
5.37a = 37.6
Make a the subject of formula
a = 37.6 / 5.37
a = 7.0 m/s^2
Therefore, the acceleration of the plane to bring the plane to a halt is 7 m/s^2
Explanation:
The five-step process for treating a muscle or joint injury such as an ankle sprain is called "P.R.I.C.E." which is short for Protection, Rest, Ice, Compression, and Elevation).
