Answer:
The thickness of the oil slick is 
Explanation:
Given that,
Index of refraction = 1.28
Wave length = 500 nm
Order m = 1
We need to calculate the thickness of oil slick
Using formula of thickness

Where, n = Index of refraction
t = thickness
= wavelength
Put the value into the formula



Hence, The thickness of the oil slick is 
Answer:
Condensation
Explanation:
That is when water from the air collects as a drop. also can be rain.
Part a)
per day electricity power consumed when 100 W bulb is used for 8 hours

for one year consumption

now the cost will be given

now when other energy efficient light is used

for one year consumption

now the cost will be given

the process by which green plants and some other organisms use sunlight to synthesize foods from carbon dioxide and water. Photosynthesis in plants generally involves the green pigment chlorophyll and generates oxygen as a byproduct.The inputs of photosynthesis are light energy, and matter in the form of water absorbed through the roots, and carbon dioxide absorbed through the leaves.The main outputs are oxygen, which is released into the air, and glucose sugar (chemical energy), which is used to keep the plant alive.
Answer:
The minimum coefficient of friction is 0.27.
Explanation:
To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.
Centripetal force is written as

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

with ω denoting the angular velocity, which we are given. With that, the above becomes:

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

and so 45 rev/min = 4.71 rad/s.

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.