Complete Question
An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are investigating the environmental effects of the accident and need to know the area of the spill. The tanker captain informs you that 18000 liters of oil have escaped and that the oil has an index of refraction of n = 1.1. The index of refraction of the ocean water is 1.33. From the deck of your ship you note that in the sunlight the oil slick appears to be blue. A spectroscope confirms that the dominant wavelength from the surface of the spill is 485 nm. Assuming a uniform thickness, what is the largest total area oil slick
Answer:
The largest total area of the oil slick 
Explanation:
From the question we are told that
The volume of oil the escaped is 
The refractive index of oil is 
The refractive index of water is 
The wavelength of the light is 
Generally the thickness of the oil for condition of constructive interference between the oil and the water is mathematically represented as

Where is the order of interference of the light and it value ranges from 1, 2, 3,...n
It is usually take as 1 unless stated otherwise by the question
substituting value
The are can be mathematically evaluated as

Substituting values


Answer:
Explanation:
As we know that the ball is projected upwards so that it will reach to maximum height of 16 m
so we have

here we know that

also we have

so we have


Now we need to find the height where its speed becomes half of initial value
so we have

now we have





Answer:
<u>Foot per second. Foot-pound-second system. Frames per second, the frequency (rate) at which consecutive images (frames) appear on a display.</u>
Explanation:
:)
Answer:
Explanation:
a ) Between r = 0 and r = r₁
Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .
b ) From r = r₁ to r = r₂
At distance r , charge contained in the sphere of radius r
volume charge density x 4/3 π r³
q = Q x r³ / R³
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q x r³ / ε₀R³
E= Q x r / (4πε₀R³)
E ∝ r .
c )
Outside of r = r₂
charge contained in the sphere of radius r = Q
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q / ε₀
E = Q / 4πε₀r²
E ∝ 1 / r² .