A) initial volume
We can calculate the initial volume of the gas by using the ideal gas law:

where

is the initial pressure of the gas

is the initial volume of the gas

is the number of moles

is the gas constant

is the initial temperature of the gas
By re-arranging this equation, we can find

:

2) Now the gas cools down to a temperature of

while the pressure is kept constant:

, so we can use again the ideal gas law to find the new volume of the gas

3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:

by using the data we found at point 1) and 2), we find

where the negative sign means the work is done by the surrounding on the gas.
You could get sick by breathing throw your mouth and you have a less chance of getting sick by breathing throw your nose.
Early hypotheses were not based on observations.
Early hypotheses were not tested by experimentation.
Early hypotheses were formed from scientific questions.
Early hypotheses were influenced by creative thinking
Answer:
4.163 m
Explanation:
Since the length of the bridge is
L = 380 m
And the bridge consists of 2 spans, the initial length of each span is

Due to the increase in temperature, the length of each span increases according to:

where
is the initial length of one span
is the temperature coefficient of thermal expansion
is the increase in temperature
Substituting,

By using Pythagorean's theorem, we can find by how much the height of each span rises due to this thermal expansion (in fact, the new length corresponds to the hypothenuse of a right triangle, in which the base is the original length of the spand, and the rise in heigth is the other side); so we find:

Answer:
Intensity of the light (first polarizer) (I₁) = 425 W/m²
Intensity of the light (second polarizer) (I₂) = 75.905 W/m²
Explanation:
Given:
Unpolarized light of intensity (I₀) = 950 W/m²
θ = 65°
Find:
a. Intensity of the light (first polarizer)
b. Intensity of the light (second polarizer)
Computation:
a. Intensity of the light (first polarizer)
Intensity of the light (first polarizer) (I₁) = I₀ / 2
Intensity of the light (first polarizer) (I₁) = 950 / 2
Intensity of the light (first polarizer) (I₁) = 425 W/m²
b. Intensity of the light (second polarizer)
Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ
Intensity of the light (second polarizer) (I₂) = (425)(0.1786)
Intensity of the light (second polarizer) (I₂) = 75.905 W/m²