UT PHY 317K - HW01-solutions (12 pages)

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HW01-solutions



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HW01-solutions

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Pages:
12
School:
University of Texas at Austin
Course:
Phy 317k - General Physics I
General Physics I Documents
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gilbert hlg574 HW01 tsoi 55730 This print out should have 28 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points The number x 0 0675 has how many significant digits Correct answer 57 1466 km h Explanation Let tother 17 min The total time is t t1 t2 t3 tother 34 8 min 11 1 min 46 3 min 17 min 1 82 h so Correct answer 3 Explanation All non zero digits are significant Any zeros in front of the first non zero digit are placeholders zeros between non zero digits are significant 1 vav 104 007 km x 57 1466 km h t 1 82 h 002 part 1 of 2 10 0 points A person travels by car from one city to another She drives for 34 8 min at 82 1 km h 11 1 min at 100 km h 46 3 min at 49 1 km h and spends 17 min along the way eating lunch and buying gas Determine the distance between the cities along this route 004 10 0 points The velocity of the transverse waves produced by an earthquake is 4 km s while that of the longitudinal waves is 7 28 km s A seismograph records the arrival of the transverse waves 51 1 s after that of the longitudinal waves How far away was the earthquake Correct answer 104 007 km Correct answer 453 668 km Explanation Explanation Let t1 v1 t2 v2 t3 v3 34 8 min 82 1 km h 11 1 min 100 km h 46 3 min and 49 1 km h x x1 x2 x3 v1 t1 v2 t2 v3 t3 82 1 km h 34 8 min 100 km h 11 1 min 49 1 km h 46 3 min 104 007 km 003 part 2 of 2 10 0 points Determine the average speed for the trip Let vt 4 km s v 7 28 km s t 51 1 s and If the distance to the earthquake is d then the time lag is d d d v vt vt v v vt t v vt d v vt 51 1 s 7 28 km s 4 km s 4 km s 7 28 km s t 453 668 km Note The longitudinal wave travels faster than the transverse waves In fact the ratio gilbert hlg574 HW01 tsoi 55730 of the longitudinal velocity to the transverse wave velocity should be about 3 9 005 part 1 of 3 10 0 points Consider the plot below describing motion along a straight line with an initial position of x0 10 m 7 5 velocity m s 7 velocity m s 6 4 8 6 5 3 2 0 bb 4 3 2 2 b 1 2 1 2 3 4 5 6 7 time s What is the position at 2 seconds b 1 2 3 4 5 time s b 6 7 8 9 b The position at 2 seconds is 10 meters plus the area of the triangle shaded in the above plot b 0 b bb 1 1 1 b 8 b 9 2 8 b 9 Correct answer 11 m 1 2 s 0 s 2 1 m s 0 m s x 10 m 11 m however it can also be calculated 1 tf ti 2 2 10 m 0 m s 2 s 0 s 1 0 5 m s2 2 s 0 s 2 2 11 m x xi vi tf ti 006 part 2 of 3 10 0 points What is the position at 6 seconds Correct answer 31 m Explanation The position is 11 m plus the area of the trapezoid from 2 s to 6 s 1 6 s 2 s 2 9 m s 1 m s x 11 m Explanation The initial position given in the problem is 10 m 31 m gilbert hlg574 HW01 tsoi 55730 3 however it can also be calculated 2 with a decreasing speed 1 x xi vi tf ti tf ti 2 2 11 m 1 m s 6 s 2 s 1 2 m s2 6 s 2 s 2 2 31 m 3 with a constant speed Explanation dv a As long as the acceleration is dt positive the velocity is always increasing 007 part 3 of 3 10 0 points What is the position at 8 seconds Correct answer 29 6667 m Explanation The position is 31 m minus the area of the triangle from 6 s to 8 s 1 8 s 6 s 2 1 33333 m s 0 m s x 31 m 009 10 0 points An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train When the engineer first sees the car the locomotive is 360 m from the crossing and its speed is 16 m s If the engineer s reaction time is 0 53 s what should be the magnitude of the minimum deceleration to avoid an accident Correct answer 0 364133 m s2 Explanation 29 6667 m however it can also be calculated Let 1 tf ti 2 2 31 m 0 m s 8 s 6 s 1 0 666667 m s2 8 s 6 s 2 2 29 6667 m d 360 m v vi 16 m s t 0 53 s x xi vi tf ti and While the engineer reacts the train moves d v t 16 m s 0 53 s 8 48 m 008 10 0 points The diagram describes the acceleration vs time behavior for a car moving in the xdirection forward so it now has to decelerate to rest within a displacement of x d d 360 m 8 48 m 351 52 m vf2 vi2 2 a x 0 a Pb vi2 16 m s 2 2 x 2 351 52 m 0 364133 m s2 Q b a which has a magnitude of 0 364133 m s2 0 At the point Q the car is moving 1 with an increasing speed correct t 010 part 1 of 2 10 0 points Consider the free fall of a stone It begins from rest at t 0 Denote the distance covered from t 0 to t t1 by s1 from t 0 and gilbert hlg574 HW01 tsoi 55730 hmax and falls back toward Earth On the way down it is caught at the same height at which it was thrown upward 2 5 b b b b b b b b v0 1 2 bbbb bb b b b b b b 3 3 hmax to t t2 by s2 the speed at t t1 and t t2 to be v1 and v2 respectively t2 s2 If 2 find the ratio t1 s1 4 4 4 correct 5 1 6 8 Explanation 1 1 s s0 v0 t a t2 g t2 2 2 since s0 0 v0 0 and a g so t2 s2 22 s1 t1 2 t2 22 4 t1 011 part 2 of 2 10 0 points v2 Find the …


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