I think you should hold a stretch for 10-30 seconds
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
Learn more about terminal velocity here:
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Answer:
The displacement is 386.16m
Explanation:
A seal dives to a depth of 248m. To find displacement, we must calculate the resultant vectors which will give us the displacement
R= sqrt(vector1+vector2)
Since this is a right angle triangle
R= sqrt(248^2 + 296^2)
R= sqrt(149120)
R= 386.16m
Displacement = 386.16m
The answer is C) The water pushed up on the skis
The water reacts to the downward force of the skis by pushing back up against the skis.
KE=1/2mv^2 - equation for kinetic energy
KE=(1/2)(0.12 kg)((7.8 m/s)^2 - plug it into the formula
KE=(0.06 kg)(60.84 m/s) - multiply 1/2 to the mass and square the speed
KE= 3.7 J - answer
Hope this helps