If the original pressure is 5 atm and original volume is 100 L, and the new volume is 20 L, what is the new pressure?
1 answer:
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Answer:
a) 2.76x10^-9 m/s^2
b) downward
Explanation:
a) In order to do this, we need to use the newton's law of gravitation which is:
F = G.m/r²
Where:
G: Constant of gravitation (6.67x10^-11 N/m)
m: mass of the object
r: distance or radius of the object.
In this case, we have two spheres, so this law becomes:
F = G*m1*m2 / r²
Now, you are not providing a picture with this, however, this problem is very similar to one I solved before, so, I'm gonna use the same values to solve this. If, you have different values for the distance, just replace them in this procedure and should get the correct and more accurate answer.
In this case, the distance that separates sphere in P from sphere A and B is 10 cm. As both spheres have the same mass and they are at the same distance from P, we can assume that F1 = F2 so:
F1 = (6.67x10^-11) * (0.01) * (0.26) / (0.1)² = 1.73x10^-11 N
sphere at P, will roll downward to the middle of sphere A and B. The distance between these two points it's 8 cm (See picture attached)
Now, according to my data, they are exerting a force in the y axis, so:
F1y = Fy * senФ
senФ using trigonometry is:
senФ = opposite lenght / hipotenuse
senФ = 8/10 = 0.8
Replacing:
F1y = 1.73x10^-11 * 0.8 = 1.38x10^-11 N
This force it's exactly the same as F2y, so the total force is:
F = 1.38x10^-11 * 2 = 2.76x10^-11 N
Finally, using the second law of Newton:
F = m*a
Solving for a:
a = F/m
replacing:
a = 2.76x10^-11 / 0.01
a = 2.76x10^-9 m/s²
According to the picture and the value of this magnitude, it's going downward.
