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Answer:
Given:
P₁ = 500 kPa
T₁ = 860 K
P₂= 100 kPa
T₂ = 460 K
Let's take entropy properties of T1 and T2 from ideal properties of air,
at T = 860K, s(T₁) = 2.79783 kJ/kg.K
at T = 460K, s(T₂) = 2.13407 kJ/kg.K
using entropy balance equation:
= - 0.2018 kJ/kg. K
In this case the entropy is negative, which means the value of exit temperature is not correct, beacause entropy should always be positive(>0).
Answer:
1. Graph C
2. Friction
Explanation:
1. The line on all of the graphs shown represents velocity. The formula for velocity is where d is distance and t is time. Focusing on the first lap, the starting point on the graph should be the origin and the "ending" point should be (20, 3). These requirements eliminate graph A as an answer because its "end" is not (20, 3). During the break, the student does not move, so the slope of the line should be completely horizontal. The break lasted for 5 minutes, so the correct graph should have a horizontal line between the points (20, 3) and (25, 3). This requirement eliminates graph B and D because their break is either not long enough (B) or too long (D).
2. Friction slows down the movement of objects. When an object is rough, it produces more friction which causes the object to be slowed more. When an object is smooth, friction slows it less than it would for a rough object.