Not sure the precise concept of "normal observation", but I assume that is observed by "eyes".
Eye observation is basically macroscopic, but when you use a mark, which can be regarded as a point of mass, then it goes to microscopic.
Mark is a reference point which you can compare the relative position change, but with your eyes, first you cannot notice microscopic changes, second the eyes cannot precisely set a stable reference point.
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:
brainly.com/question/20409472
#SPJ4
Answer:
if this surface has a higher index than in the medium where the light travels, the reflected wave has a phase change of 180º
Explanation:
When a ray of light falls on a surface if this surface has a higher index than in the medium where the light travels, the reflected wave has a phase change of 180º this can be explained by Newton's third law, the light when arriving pushes the atoms of the medium that is more dense, and these atoms respond with a force of equal magnitude, but in the opposite direction.
When the fractional index is lower than that of the medium where the reflacted beam travels, notice a change in phase.
Also, when light penetrates the medium, it modifies its wavelength
λ = λ₀ / n
We take these two aspects into account, the condition for contributory interference is
d sin θ = (m + 1/2) λ
for destructive interference we have
d sin θ = m λ
in general this phenomenon is observed at 90º
2 d = (m +1/2) λ° / n
2nd = (m + ½) λ₀
Answer:
First, let’s correct the question. Acceleration is the rate of change in velocity. Its unit therefore is ft/sec/sec. If S is the distance traveled for a given duration, S = Vot + (1/2)at^2 where Vo is the initial velocity, a is the acceleration and t is the time. For Vo = 0, a = 6m/sec/sec and t = 3 sec. The distance traveled is S = 0 + (1/2) x 6 x 3^2 = 27 meters
To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by
Where,
T = Tension
Linear density
Our data are given by
Tension , T = 70 N
Linear density , 
Amplitude , A = 7 cm = 0.07 m
Period , t = 0.35 s
Replacing our values,
Speed can also be expressed as
Re-arrange to find \lambda
Where,
f = Frequency,
Which is also described in function of the Period as,
Therefore replacing to find 
Therefore the wavelength of the waves created in the string is 3.49m
