practice test 3 - brown (twb493) midterm 03 practice turner...

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Unformatted text preview: brown (twb493) midterm 03 practice turner (57340) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A 300 g block of aluminum at a tempera- ture of 500 K is placed in intimate contact with a 600 g block of iron at 300 K. The blocks are contained within an insulated en- closure. What is the final temperature of the two blocks? Given: The specific heat capac- ity of aluminum is 1 J / K / g and the specific heat capacity of iron is 0 . 4 J / K / g. Correct answer: 411 . 1 K. Explanation: Let : T Al = 500 K , m Al = 300 g , C Al = 1 J / K / g , T Fe = 300 K , m Fe = 600 g , and C Fe = 0 . 4 J / K / g . Let the final temperature be T f . Heat will flow from the hot object to the cold object subject to the constraint that the net thermal energy will be conserved. E thermal = C m T E thermal ( Al ) + E thermal ( Fe ) = 0 C Al m Al T Al + C Fe m Fe T Fe = 0 (1)(600)( T f- 500 K) +(0 . 4)(300)( T f- 300 K) = 0 ( T f- 500 K) + 0 . 8( T f- 300 K) = 0 Thus, the final temperature T f is T f = 411 . 1 K 002 10.0 points A metal block of mass 3 kg is falling down- ward with a velocity of 1 . 6 m / s. This block is initially 0 . 8 m above the floor. When the block is 0.4 m/s, it strikes the top of a re- laxed vertical spring of length 0.4 m. The spring constant is 2000 N/m. After strik- ing the spring, the block rebounds. What is the maximum height above the floor that the block reaches after the impact? Given that the value of g is 9 . 8 m / s 2 Correct answer: 0 . 9306 m. Explanation: Let : m = 3 kg , v i = 1 . 6 m / s , y i = 0 . 8 m , and g = 9 . 8 m / s 2 . This is a cute problem. Since the poten- tials involved are all conservative, i.e., they only depend on distance, the existence of the spring is not relevant save to gently reverse the motion. We can think of the block having its velocity reversed and ask how high it will go. E i = mgy i + 1 2 mv 2 i = E f = mgy f + 1 2 mv 2 f We set v f = 0 and then cancel out the masses on both sides of the equation which leaves us with y f = y i + v 2 i 2 g y f = 0 . 8 m + (1 . 6 m / s) 2 2(9 . 8 m / s 2 ) = 0 . 9306 m 003 (part 1 of 3) 10.0 points brown (twb493) midterm 03 practice turner (57340) 2 Given the potential energy curve to the in- teraction of two atoms as shown below. The system is in a vibrational state indicated by the heavy horizontal line. Answer the follow- ing questions: (a) At r = r 1 , what is the approximate value of the potential energy of the molecule? 1.-1.0 eV 2.-0.9 eV 3.-1.4 eV 4.-0.2 eV 5.-0.7 eV 6.-1.6 eV 7.-0.4 eV 8.-1.2 eV 9.-1.3 eV correct Explanation: (a) The potential energy is about -1.3 eV....
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This note was uploaded on 04/13/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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practice test 3 - brown (twb493) midterm 03 practice turner...

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