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Engines and Review. Exam 1 Review Material You drive 8.0 km at 40 km/h and then another 5.0 km at 90 km/h. Your average speed over the drive will be….

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Presentation on theme: "Engines and Review. Exam 1 Review Material You drive 8.0 km at 40 km/h and then another 5.0 km at 90 km/h. Your average speed over the drive will be…."— Presentation transcript:

1 Engines and Review

2 Exam 1 Review Material You drive 8.0 km at 40 km/h and then another 5.0 km at 90 km/h. Your average speed over the drive will be….

3 Solution V= d/t T= d/v = 8/ 40 =.2 T2 = 5/ 90 =.0555 T total =.25555 D total = 8 + 5 =13 V total = 13 /.2555 = 50. 87 km / h

4 Problem 2: If vector has components Ax = −5.0 lb and Ay = −2.0 lb, and vector has components Bx = 5.0 lb and By = −10.0 lb, what is the magnitude of vector C ?

5 Solution: Cx = -5 – 5 = -10 Cy = -2 - - 10 = 8 C total = ( 10^ 2 + 8 ^2 ) ^ ½ C total = 12.8 lbs

6 Problem 3: A ball is thrown with an initial velocity of 60 m/s at an angle of 20° above the horizontal. If we can neglect air resistance, what is the horizontal component of its instantaneous velocity at the exact top of its trajectory?

7 Solution: Vx = F cos degree Vx = 60 cos 20 = 24.5 m/s

8 Question: An object slides on a level floor. It slows and comes to a stop with a constant acceleration of magnitude 3.5 m/s2. What is the coefficient of kinetic friction between the object and the floor?

9 Solution: F=ma U k = a/g = 3.5 / 9.8 =.36

10 Last Review Question: A 25.0-kg crate is being pulled along a horizontal smooth surface. The pulling force is 20.0 N and is directed at an angle 20.0° above the horizontal. What is the magnitude of the acceleration of the crate?

11 Solution: Ax = Fx /m Fx= F cos degree = 20 cos 20 Fx = 8.16 N Ax= 8.16 / 25 =.32 m/s^2

12 Today’s Problems: Carnot engine operates between hot and cold reservoirs with temperatures 600°C and −25.0 °C, respectively. If the engine performs 2300.0 J of work per cycle, how much heat is extracted per cycle from the hot reservoir?

13 Solution: e = 1 – Q cold / Q heat E= 1- Tc/Th W= e Qh Qh = W /e = W / (1- Tc/ T h )  2300 / ( 1- 873.15 / 248.15 ) Qh = 913.192 ( absolute value )

14 Last Problem: 1200 J of heat spontaneously flowing through a copper rod from a hot reservoir at 650 K to a cold reservoir at 350 K. Determine the amount by which this process changes the entropy of the universe.

15 Solution: 1600 J of heat spontaneously flowing through a copper rod from a hot reservoir at723K to a cold reservoir at 500 K. Determine the amount by which this process changes the entropy of the universe.

16 Solution: S = (Qc / Tc ) – ( Q h – T h) S=( 1200/ 500) – (1200 / 723) S=.74

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