It decreases because it gave its momentum to the other car.
It can be described as a constant variation
Answer:
a. Ssystem > 40 J/K
Explanation:
Given that
The entropy of first block = 10 J/K
The entropy of second block = 30 J/K
When two bodies come into contact with each other, the entropy of the combined system will increase and the entropy sum remains unchanged: According to the Second law of thermodynamics.The entropy of the system will be greater than 40 J/K.
Therefore the answer is a.
Ssystem > 40 J/K
By definition we have that the final speed is:
Vf² = Vo² + 2 * a * d
Where,
Vo: Final speed
a: acceleration
d: distance.
We cleared this expression the acceleration:
a = (Vf²-Vo²) / (2 * d)
Substituting the values:
a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
a = -77268 mi / h ^ 2
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
First you must make a free body diagram and see the acceleration of the car:
g = 32.2 feet / sec ^ 2
a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
a = -31.48 feet / sec ^ 2
A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
A = -22.07 feet / sec ^ 2
Clearing the braking distance:
Vf² = Vo² + 2 * a * d
d = (Vf²-Vo²) / (2 * a)
Substituting the values:
d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
d = 175.44 feet
answer:
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
Since the tower base is square with a side length of 125 m,
Therefore,
Square root of 31250 = 176.776953 (Diameter)
, so this is the diameter of the cylinder to enclose it, and radius, r = 88.38834765 m and height, h = 324 m.
The volume of cylinder,
Thus, the mass of the air in the cylinder,
Hence, the mass of the air in the cylinder is this more than the mass of the tower.
