Answer:
12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor are equivalent to 6 drinks.
Explanation:
In the United States, a standard "drink" of beer has 12 ounces, a standard "drink" of wine has 5 ounces and standard drink of liquor has 1.5 ounces. Then, we obtain the quantity of drinks by dividing the total volume of each drink by its respective unit volume and summing each term. That is:
12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor are equivalent to 6 drinks.
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
5.4 x 1014Hz
wavelength x frequency = the speed of light
The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is
<em>T</em> - 25 N = (8 kg) <em>a</em>
where <em>T</em> is the tension in the rope and <em>a</em> is the acceleration.
The hanging mass has a net force of
(6 kg) <em>g</em> - <em>T</em> = (6 kg) <em>a</em>
where <em>g</em> = 9.8 m/s².
Adding these equations together eliminates <em>T</em>, and we can solve for <em>a</em> :
(<em>T</em> - 25 N) + ((6 kg) <em>g</em> - <em>T </em>) = (14 kg) <em>a</em>
33.8 N = (14 kg) <em>a</em>
<em>a</em> = (33.8 N) / (14 kg) ≈ 2.4 m/s²
Then the tension in the rope is
<em>T</em> - 25 N = (8 kg) (2.4 m/s²)
<em>T</em> ≈ 25 N + 19.31 N ≈ 44 N