gilbert (hlg574) – HW01 – tsoi – (55730) 1This print-out should have 28 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsThe number x = 0.0675 has how many signif-icant digits?Correct answer: 3.Explanation:All non-zero digits are significant. Anyzeros in front of the first non-zero digit areplaceholders; zeros between non-zero digitsare significant.002 (part 1 of 2) 10.0 pointsA person travels by car from one city to an-other. She drives for 34.8 mi n at 82.1 km/h,11.1 min at 100 km/h, 4 6.3 min at 49.1 km/h,and sp ends 17 min along the way eating lunchand buying gas.Determine the distance between the citiesalong this route.Correct answer: 104.007 km.Explanation:Let : t1= 34.8 min ,v1= 82.1 km/h ,t2= 11.1 min ,v2= 100 km/h ,t3= 46.3 min , andv3= 49.1 km/h .x = x1+ x2+ x3= v1t1+ v2t2+ v3t3= (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 pointsDetermine the average speed for the trip.Correct answer: 57.1466 km/h.Explanation:Let : tother= 17 min .The total time ist = t1+ t2+ t3+ tother= 34.8 min + 11.1 min+ 46.3 min + 17 min= 1.82 h , sovav=xt=104.007 km1.82 h=57.1466 km/h .004 10.0 pointsThe velocity of the transverse waves producedby an earthquake is 4 km/s, while that of thelongitudinal waves is 7.28 km/s. A seismo-graph records the arrival of the transversewaves 51.1 s after that of the longitudinalwaves.How far away was the earthquake?Correct answer: 453.668 km.Explanation:Let : vt= 4 km/ s ,vℓ= 7.28 km/s , andt = 51. 1 s .If the distance to the earthquake is d, thenthe time lag is∆t =dvt−dvℓ=d (vℓ− vt)vℓvtd =∆t vℓvtvℓ− vt=(51.1 s) (7.28 km/s) (4 km/s)4 km/s − 7.28 km/s=453.668 km .Note: The longitudinal wave travels fasterthan the transverse waves. In fact, the ratiogilbert (hlg574) – HW01 – tsoi – (55730) 2of the l ongitudinal velocity to the transversewave velocity should be about√3 .005 (part 1 of 3) 10.0 pointsConsider the plot below describing motionalong a straight line with an initial position ofx0= 10 m.−2−101234567891 2 3 4 5 6 7 8 9bbbbbtime (s)velocity (m/s)What is t he position at 2 seconds?Correct answer: 11 m.Explanation:The initial position gi ven in the problem is10 m.bbbbbbbb1 2 3 4 5 6 7 8 90123456789−1−2time (s)velocity (m/s)The position at 2 seconds is 10 meters pl usthe area of the triangle (shaded in the aboveplot)x = 10 m +12(2 s − 0 s)× (1 m/s − 0 m/s)=11 m ;however, i t can also be calculated:x = xi+ vi(tf− ti) +12(tf− ti)2= (10 m) + (0 m/s) (2 s − 0 s)+12(0.5 m/s2) (2 s − 0 s)2= 11 m .006 (part 2 of 3) 10.0 pointsWhat is the position at 6 seconds?Correct answer: 31 m.Explanation:The position is 11 m plus the area of thetrapezoid from 2 s to 6 sx = 11 m +12(6 s − 2 s)× (9 m/s + 1 m/s)=31 m ;gilbert (hlg574) – HW01 – tsoi – (55730) 3however, i t can also be calculated:x = xi+ vi(tf− ti) +12(tf− ti)2= (11 m) + (1 m/s) (6 s − 2 s)+12(2 m/s2) (6 s − 2 s)2= 31 m .007 (part 3 of 3) 10.0 pointsWhat is the position at 8 seconds?Correct answer: 29.6667 m.Explanation:The position is 31 m minus the area of thetriangle from 6 s to 8 sx = (31 m) +12(8 s − 6 s)× (−1.333 3 3 m/s − 0 m/s)=29.6667 m ;however, i t can also be calculatedx = xi+ vi(tf− ti) +12(tf− ti)2= 31 m + (0 m/ s) ( 8 s − 6 s)+12(−0.666667 m/s2) (8 s − 6 s)2= 29.6667 m .008 10.0 pointsThe diagram describes the acceleration vstime behavior for a car moving i n t he x-direction.b baPQt0At the point Q, the car is moving1. with an increasing speed. correct2. with a decreasing speed.3. with a constant speed.Explanation:a =d vdt. As long as the acceleration ispositive the velocity is always increasing.009 10.0 pointsAn engineer in a locomotive sees a car stuckon the track at a railroad crossing in front ofthe train. When t he engineer first sees thecar, the locomotive is 360 m from the crossingand its speed is 16 m/s.If the engi neer’s reaction time is 0. 53 s,what should be the magnitude of the mini-mum deceleration to avoid an accident?Correct answer: 0.364133 m/s2.Explanation:Let : d = 360 m ,v = vi= 16 m/s , andt = 0.5 3 s .While the engineer reacts, the t rain moves∆d = v t = (16 m/s) (0.53 s) = 8.48 mforward, so it now has to decelerate to restwithin a displacement of∆x = d−∆d = 360 m−8.48 m = 351.52 m , andv2f= v2i+ 2 a ∆x = 0a =−v2i2 ∆x=−(16 m/s)22 (351.52 m)= −0.364133 m/s2,which has a magnitude of 0.364133 m/s2.010 (part 1 of 2) 10.0 pointsConsider the free fall of a stone. It beginsfrom rest at t = 0. Denote the distancecovered from t = 0 to t = t1by s1, from t = 0gilbert (hlg574) – HW01 – tsoi – (55730) 4to t = t2by s2, the speed at t = t1and t = t2to be v1and v2, respectively.Ift2t1= 2, find t he ratios2s1.1. 22. 53. 34. 4 correct5. 16. 8Explanation:s = s0+ v0t +12a t2=12g t2since s0= 0, v0= 0 and a = g, sos2s1=t22t21=t2t12= 22= 4 .011 (part 2 of 2) 10.0 pointsFind the ratiov2v1.1. 52. 13. 2 correct4. 45. 86. 3Explanation:Similarly,v = v0+ a t = g t , sov2v1=t2t1= 2 .012 (part 1 of 2) 10.0 pointsA ball is thrown upward with an initial ver-tical velocity of v0to a m a ximum height ofhmaxand falls back toward Earth. On t heway down, it is caught at the same height atwhich it was thrown upward.bbbbbbbbbbbbbbbbbbbbv0hmaxIf the total time the ball remains in the airis t, find its speed when caught. The acceler-ation of gravity is g. Neglect air resistance.1. vf=√2 g t2. vf=14g t3. vf= 2 g t4. vf= g t5. vf=1√2g t6. vf=12g tcorrect7. vf= 4 g tExplanation:With up positi ve, a = −g and vf= 0 forthe upward motion, sovf= v0+ a tup0 = v0− g tuptup=v0g.From symmetry, tup= tdown, and thespeeds are the same at the same height, sot = 2 tdown=2 vfgvf=12g t .gilbert (hlg574) – HW01 – tsoi – (55730) 5013 (part 2 of 2) 10.0 pointsIf the time the ball remains in the air is t, findthe maximum height the ball at tained whilein the air.1. hmax= 8 g t22. hmax= 2 g t23. hmax= 4 g t24. hmax=12g t25. hmax=14g t26. hmax=18g t2correct7. hmax= g t2Explanation:The velocity at the top is zero. For uppositive, a = −g and v0= g tup, sohmax= v0tup−12g t2up= g t2up−12g t2up=12g t2uptup=s2 hmaxg.From sy m met ry, tdown= tup, sottotal= 2s2 hmaxghmax=18g t2total.014 (part 1 of 2) 10.0 pointsThrow a ball upward from point O with aninitial speed of 84 m/s.thAytBhBtAbbbbbbbbbbbb bbbbbbbbbbbb84 m/sOBAWhat is the maximum height? The accel-eration of gravity