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To investigate absorption, Hector cuts two rectangles from five different-colored sheets of construction paper. He then staples three sides of each pair of rectangles to make a pocket and inserts a thermometer into each.
He takes the thermometers in the construction paper pockets outside, where it is clear and sunny. The initial temperature of each thermometer is 22°C. The image shows the thermometers after being outside for 15 minutes.
A thermometer in black paper reads 38 degrees Celsius, a thermometer in white paper reads 24 degrees Celsius, a thermometer in green paper reads 29 degrees Celsius, a thermometer in red paper reads 31 degrees Celsius, and a thermometer in blue paper reads 33 degrees Celsius.
Complete Hector's table by recording the final temperature and calculating the temperature change for each thermometer.
To investigate absorption, Hector cuts two rectangles from five different-colored sheets of construction paper. He then staples three sides of each pair of rectangles to make a pocket and inserts a thermometer into each.
He takes the thermometers in the construction paper pockets outside, where it is clear and sunny. The initial temperature of each thermometer is 22°C. The image shows the thermometers after being outside for 15 minutes.
A thermometer in black paper reads 38 degrees Celsius, a thermometer in white paper reads 24 degrees Celsius, a thermometer in green paper reads 29 degrees Celsius, a thermometer in red paper reads 31 degrees Celsius, and a thermometer in blue paper reads 33 degrees Celsius.
Complete Hector's table by recording the final temperature and calculating the temperature change for each thermometer.
1 answer:
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Answer:
0 Joules
Explanation:
The work done is given by
where,
F = Force applied
s = Displacement of the object = 0 m
= Angle between the force applied and the horizontal = 0
Work is only observed when there is a displacement.
The work done by me is 0 Joules as I was unable to move it.
Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.
From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as
We are given the second smallest nonzero thickness at which destructive interference occurs.
This corresponds to, m = 2, therefore
The index of refraction of soap is given, then
Combining the results of all steps we get
Rearranging, we find