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….

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

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 ?

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

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?

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

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?

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

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?

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

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

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

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.

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.

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