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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A. The vector goes from (4,0) to (3-2)
(x,y)
Answer:
The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.
Explanation:
Given that,
Mass = 2.15 kg
Distance = 0.0895 m
Amplitude = 0.0235 m
We need to calculate the spring constant
Using newton's second law

Where, f = restoring force


Put the value into the formula


We need to calculate the kinetic energy of the mass
Using formula of kinetic energy

Here, 

Here, 


Put the value into the formula


Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.
We know,
V= f× wavelength
10.5= f×0.15
f=10.5/0.15
f= 70 Hz
You should just ask the wave