Which ocean borders the west coast of this United States

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 inv

PolarNik [594]

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 $A = 8.257 *10^{9} \ m^2$

Explanation:

From the question we are told that

The volume of oil the escaped is $V = 18000 \ L$

The refractive index of oil is $n_o = 1.1$

The refractive index of water is $n_w = 1.33$

The wavelength of the light is $\lambda = 485 \ nm = 485 * 10^{-9} \ m$

Generally the thickness of the oil for condition of constructive interference between the oil and the water is mathematically represented as

$d = m *\frac{\lambda}{2n_w}$

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

$d = 1 * \frac{485 *10^{-9}}{2 * 1.1}$

$d = 218 nm$

The are can be mathematically evaluated as

$A = \frac{V}{d}$

Substituting values

$A = \frac{18000}{218*10^{-8}}$

$A = 8.257 *10^{9} \ m^2$

6 0

2 years ago

The drag force on a falling coffee filter can be modeled as a linear function with respect to velocity, F⃗ D =−bv, where b is a

nordsb [41]

Newton's second law and graphical analysis allow us to find the correct answer for the measure of the constant b is:

D) Stack several filters to change the mass measure the acceleration of the system.

Newton's second law establishes a relationship between the net force, the mass and the acceleration of the bodies.

∑ F = m a

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the body.

They indicate that the drag force on the filters is

F = - b v

Where b is a positive constant that depends on the shape and area of the filter and v is the speed of the filter.

Let's write Newton's second law.

W - fr = ma

W -bv = ma

In the experiment the students indicate that they can measure the position, velocity and acceleration of the body.

Based on the above, if we place several filter weights and measure their speed and acceleration in each case, we can make a graph of velocity versus acceleration, we can take the value of the constant b from the slope.

Let's analyze the different answers:

A) False. The constant b depends on the shape of the filter therefore it must be kept constant.

B) False. The constant depends on the area, so it must be kept constant.

C) May be. In this case, since we have the terminal velocity, the acceleration is zero, Newton's second law remains.

B v - W = 0

b = $\frac{W}{v}$

The problem with the method is the difficulty of measuring the compression terminal velocity.

D) True. According to the discussion of the velocity versus acceleration. graph, the constant b is equal to the slope of the graph.

E) May be. The problem with this method is finding a reliable value for the terminal velocity.

In conclusion, using Newton's second law and graphical analysis we can find the correct answer for the measure of the constant b is:

D) Stack several filters to change the mass measure the acceleration of the system.

Learn more here: brainly.com/question/2441565

6 0

2 years ago

What is the amount of heat required to raise the temperature of 200.0 g of aluminum by 10 c?

mash [69]

Q = 175.8J = 1.8• 10 2 J

5 0

3 years ago

A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the

Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge $15\times10^-6$ C is placed which is the origin.

Let B be the point where the charge $9\times 10^-6$ C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, $\frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2}$ ,