Unusual precipitation patterns
Answer:
Explanation:
Given
1 mole of perfect, monoatomic gas
initial Temperature


Work done in iso-thermal process
=initial pressure
=Final Pressure

Since it is a iso-thermal process therefore q=w
Therefore q=39.64 J
(b)if the gas expands by the same amount again isotherm-ally and irreversibly
work done is





Answer:
A=1
B=-2
Explanation:
Part A and B of the question wasn't given, however, I attached the relevant parts to solve this question as follows.
From part B as attached, it shows that the right option is C which is
2A+3B=-4
Substituting B with 3A-5 then we form the second equation as shown
2A+3(3A-5)=-4
By simplifying the above equation, we obtain
2A+9A-15=-4
Re-arranging, then
11A=-4+15
Finally
11A=11
A=1
To obtain B, we already know that 3A-5 so substituting the value of A into the above then we obtain
B=3(1)-5=-2
Therefore, required values are 1 and -2
Answer:
0.0312J
Explanation:
Let x be the distance the staple moves:

And spring constant is 

Hence, the potential energy is 0.0312J
I'm guessing that you mean like this:
-- The ruler is held with zero at the bottom, and the centimeter markings
increase as you go up the ruler.
-- You place your fingers with the ruler and the zero mark between them.
-- The number where you catch the ruler is the distance it has fallen.
Then, all we have to find is the time it takes for the ruler to fall 11.3 cm .
Here's the formula for the distance an object falls from rest
in a certain time:
Distance = (1/2) (gravity) (time)²
On Earth, the acceleration of gravity is 9.8 m/s².
So we can write ...
11.2 cm = (1/2) (9.8 m/s²) (time)²
or
0.112 meter = (4.9 m/s²) (time)²
Divide each side
by 4.9 m/s² : (0.112 m) / (4.9 m/s²) = time²
(0.112 / 4.9) sec² = time²
Square root
each side: time = √(0.112/4.9 sec²)
= √ 0.5488 sec²
= 0.74 second (rounded)