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공지사항

최근글

최신대학물리학 5판 솔루션입니다.

2

CHAPTER OUTLINE 2 Position, Velocity, and Speed 2 Instantaneous Velocity and Speed 2 Acceleration 2 Motion Diagrams 2 One-Dimensional Motion with Constant Acceleration 2 Freely Falling Objects 2 Kinematic Equations Derived from Calculus

Motion in One Dimension

ANSWERS TO QUESTIONS

Q2 If I count 5 s between lightning and thunder, the sound has

traveled bg 331 msaf5 0.. s=1 7 km. The transit time for the light

is smaller by

300 10 331

####### 906 10

8 . × ms=×. 5 ms

times,

so it is negligible in comparison.

Q2 Yes. Yes, if the particle winds up in the + x region at the end.

Q2 Zero.

Q2 Yes. Yes.

Q2 No. Consider a sprinter running a straight-line race. His average velocity would simply be the length of the race divided by the time it took for him to complete the race. If he stops along the way to tie his shoe, then his instantaneous velocity at that point would be zero.

Q2 We assume the object moves along a straight line. If its average velocity is zero, then the displacement must be zero over the time interval, according to Equation 2. The object might be stationary throughout the interval. If it is moving to the right at first, it must later move to the left to return to its starting point. Its velocity must be zero as it turns around. The graph of the motion shown to the right represents such motion, as the initial and final positions are the same. In an x vs. t graph, the instantaneous velocity at any time t is the slope of the curve at that point. At t 0 in the graph, the slope of the curve is zero, and thus the instantaneous velocity at that time is also zero.

x

t 0 t

FIG. Q2.

Q2 Yes. If the velocity of the particle is nonzero, the particle is in motion. If the acceleration is zero, the velocity of the particle is unchanging, or is a constant.

####### 21

22 Motion in One Dimension

Q2 Yes. If you drop a doughnut from rest af v = 0 , then its acceleration is not zero. A common

misconception is that immediately after the doughnut is released, both the velocity and acceleration are zero. If the acceleration were zero, then the velocity would not change, leaving the doughnut floating at rest in mid-air.

Q2 No: Car A might have greater acceleration than B, but they might both have zero acceleration, or otherwise equal accelerations; or the driver of B might have tramped hard on the gas pedal in the recent past.

Q2 Yes. Consider throwing a ball straight up. As the ball goes up, its

velocity is upward af v > 0 , and its acceleration is directed down

af a < 0. A graph of v vs. t for this situation would look like the figure to the right. The acceleration is the slope of a v vs. t graph, and is always negative in this case, even when the velocity is positive. v t v 0 ####### FIG. Q2. Q2 (a) Accelerating East (b) Braking East (c) Cruising East (d) Braking West (e) Accelerating West (f) Cruising West (g) Stopped but starting to move East (h) Stopped but starting to move West Q2 No. Constant acceleration only. Yes. Zero is a constant. Q2 The position does depend on the origin of the coordinate system. Assume that the cliff is 20 m tall, and that the stone reaches a maximum height of 10 m above the top of the cliff. If the origin is taken as the top of the cliff, then the maximum height reached by the stone would be 10 m. If the origin is taken as the bottom of the cliff, then the maximum height would be 30 m. The velocity is independent of the origin. Since the change in position is used to calculate the instantaneous velocity in Equation 2, the choice of origin is arbitrary. Q2 Once the objects leave the hand, both are in free fall, and both experience the same downward acceleration equal to the free-fall acceleration, – g. Q2 They are the same. After the first ball reaches its apex and falls back downward past the student, it will have a downward velocity equal to vi. This velocity is the same as the velocity of the second ball, so after they fall through equal heights their impact speeds will also be the same. Q2 With hg = 1 t 2 ####### 2 , (a) 05 ####### 1 ####### 2 .. hg t = a 0707 f 2. The time is later than 0 t. (b) The distance fallen is 025 ####### 1 ####### 2 .. hgt = af 05 2. The elevation is 0 h , greater than 0 h. 24 Motion in One Dimension P2 (a) Let d represent the distance between A and B. Let t 1 be the time for which the walker has the higher speed in 500 1 . ms= d t . Let t 2 represent the longer time for the return trip in −= 300 − 2 . ms d t . Then the times are t d 1 =bg 500. ms and t d 2 =bg 300. ms. The average speed is: v dd d v == dd d ####### + ####### + ####### = ####### == Total distance Total time ms ms ms ms ms ms ms 22 50030022 800 15 0 ####### 2 ####### 2150 ####### 800 ####### 375 .. . . . . ####### . bgbg bg ej ej (b) She starts and finishes at the same point A. With total displacement = 0, average velocity = 0. Section 2 Instantaneous Velocity and Speed P2 (a) At any time, t , the position is given by xt =ej 300. ms 22. Thus, at ti =300. s: xi ==ej3 00... ms 2 a3 00 sf 2 27 0 m. (b) At ttf =+300. s∆: xtf =+ej 300 .. ms 2 af 300 s∆ 2 , or xtf =+ + 270 . 180 s∆∆ej 300. m m ms 2 af t 2. (c) The instantaneous velocity at t =300. s is: v xx t t t fi t ####### = F − HG I KJ ####### =+ = →→ lim lim... ∆∆∆ ####### ∆ 00 ej18 0 msej3 00 ms 2 18 0 ms. P2 (a) at ti =15. s, xi =80. m (Point A) at tf =40. s, xf =20. m (Point B) v xx tt fi fi ####### = ####### − ####### − ####### = ####### − ####### − ####### =− = − ####### 20 80 ####### 415 ####### 60 ####### 24 ####### .. ####### . ####### . ####### . af af m s m 2 s ms (b) The slope of the tangent line is found from points C and D. bg txCC == 10 95.. s, and bg txDD == 35 0. s, , m v ≅−38. ms. FIG. P2. (c) The velocity is zero when x is a minimum. This is at t ≅ 4 s. Chapter 2 25 P2 (a) (b) At t =50. s , the slope is v ≅≅ ####### 58 ####### 23 m 2 s ms. At t =40. s, the slope is v ≅≅ ####### 54 ####### 18 m 3 s ms. At t =30. s , the slope is v ≅≅ 49 m 14 3 s ms. At t =20. s , the slope is v ≅≅ 36 m 9 4 s .0 m s. (c) a v t ####### =≅ ≅ ####### ∆ ####### ∆ ####### 23 ####### 50 ####### 46 ms s ms 2 . ####### . (d) Initial velocity of the car was zero. P2 (a) v = ####### ()− ####### ()− ####### = ####### 50 ####### 10 ####### 5 m s ms (b) v = ####### ()− ####### ()− ####### = − ####### 510 ####### 42 ####### 25 m s . ms (c) v = ####### ()− ####### ()− ####### = ####### 55 ####### 54 ####### 0 m m s s (d) v = ####### −−() ####### ()− ####### =+ ####### 05 ####### 87 ####### 5 m s s ms FIG. P2. *P2 Once it resumes the race, the hare will run for a time of t xx v fi x ####### = ####### − ####### = ####### − ####### = ####### 1000 ####### 25 m800 m 8 m s s. In this time, the tortoise can crawl a distance xxfi − =(af 02 25 )=500.. ms. s m Chapter 2 27 (b) Assuming that vvf = 4 and recognizing that vi =0 , the average acceleration during the race was a vvfi ####### − ####### = − ####### ()+++ ####### = total elapsed time ft s s 57 4 0 ft s 2 25 2 24 0 23 8 23 0 ####### . 0598 ####### .... ####### .. P2 (a) Acceleration is the slope of the graph of v vs t. For 0<< t 5 00. s, a =0. For 15 0 << t 20 0 .. s s, a = 0. For 50 << t 150.. s s, a vv tt fi fi ####### = ####### − ####### − ####### . a = ####### −−() ####### − ####### = ####### 800 800 ####### 15 0 5 00 ####### 160 ####### .. ####### .. . ms 2 We can plot at () as shown. 015 10 520 t (s) a (m/s 2 ) ####### FIG. P2. (b) a vv tt fi fi ####### = ####### − ####### − (i) For 5 00.. s s<< t 15 0 , ti =500. s, vi =−800. ms, t v a vv tt f f fi fi ####### = ####### = ####### = ####### − ####### − ####### = ####### −− ####### − ####### = ####### 15 0 ####### 800 ####### 800 800 ####### 15 0 5 00 ####### 160 ####### . ####### . ####### .. ####### .. ####### .. s ms ms 2 a f (ii) ti =0, vi =−800. ms, tf =20 0. s, vf =800. ms a vv tt fi fi ####### = ####### − ####### − ####### = ####### −−() ####### − ####### = ####### 800 800 ####### 20 0 0 ####### 0800 ####### .. ####### . . ms 2 P2 xt =+200 300.. ,− t 2 v dx dt ==300 200..− t, a dv dt ####### ==−200. At t =300. s: (a) x =+()200 900 900− m m=200.... (b) v =()300 600− ms=−300.. ms. (c) a = −200. ms 2 28 Motion in One Dimension P2 (a) At t =200. s, x =() 300200 .. 2 − 200200 ..()+ = 300. m m 110 .. At t =300. s, x =−+ 300900 . 2 200300 300.. .af m m= 240. so v x t ####### == − ####### − ####### ∆ = ####### ∆ ####### 24 0 11 0 ####### 200 ####### . 0 ####### . m m. 3 s s ms. (b) At all times the instantaneous velocity is v d dt = ch 300 ..... tt 2 − 200 += 300 () 600 t − 200 ms At t =200. s, v =()600 200− 200 ms= 100... ms.. At t =300. s, v =()600 300− 200 ms= 160.. ms... (c) a v t ####### == ####### − ####### − ####### ∆ = ####### ∆ ####### 16 0 10 0 ####### 300 200 ####### 600 ####### .. ####### .. ####### . ms ms s s ms 2 (d) At all times a d dt = ()600 200− =600... ms 2. (This includes both t =200. s and t =300. s). P2 (a) a v t ####### == =∆ ####### ∆ ####### 800 ####### 600 ####### 13 ####### . ####### . ####### . ms s ms 2 (b) Maximum positive acceleration is at t =3 s , and is approximately 2 m s 2. (c) a = 0 , at t =6 s, and also for t >10 s.

(d) Maximum negative acceleration is at t =8 s , and is approximately −15. ms 2.

Section 2 Motion Diagrams

P2 (a)

(b)

(c)

(d)

(e)

continued on next page

30 Motion in One Dimension

*P2 (a) Choose the initial point where the pilot reduces the throttle and the final point where the boat passes the buoy:

xi =0, xf =100 m , vxi =30 m s , vxf =?, ax =−35. ms 2 , t =?

xxvt atfix =+ + i x

####### 1

####### 2

####### 2 :

####### 100 0 30 1

####### 2

m m=+ +a s mf tt ch− 35. s 22

ch 175. ms 2 tt 2 −af 30 ms+= 100 m 0.

We use the quadratic formula:

t bb ac a

####### =− ± −

####### 24

####### 2

t =

####### ± − ()

####### =

####### ±

####### =

####### 30 900 4 1 75 100

####### 2175

####### 30 14 1

####### 35

####### 12 6

ms m s ms m ms

ms ms ms

s

22 2 2 2

####### .

####### .

####### .

####### .

####### .

ch

ch

or 453. s.

The smaller value is the physical answer. If the boat kept moving with the same acceleration, it would stop and move backward, then gain speed, and pass the buoy again at 12 s.

(b) vvatxf =+= − xi x 30 msej3 5… m s 2 4 53 s=14 1 ms

P2 (a) Total displacement = area under the af vt , curve from t = 0

to 50 s.

####### ∆

####### ∆

x

x

####### =+−

####### +

####### =

####### 1

####### 2

####### 50 15 50 40 15

####### 1

####### 2

####### 50 10

####### 1875

ms s ms s

ms s

m

bga f bga f

bgaf

(b) From t =10 s to t =40 s , displacement is

∆ x =+ + = 1 2

bg 50 ms 33 ms saf 5 bg 50 msaf 25 s 1 457 m.

####### FIG. P2.

(c) 01 ≤≤ t 5 s: a v 1 t

####### 50 0

####### 15 0

####### == 33

####### ()−

####### −

####### ∆ =

####### ∆

ms s

. ms 2 15 s s<< t 40 : a 2 = 0 40 s s≤≤ t 50 : a v 3 t ####### 050 ####### 50 40 ####### == 50 ####### ()− ####### − ####### ∆ = − ####### ∆ ms s s . ms 2 continued on next page Chapter 2 31 (d) (i) xa 110 1 t 22 t 2 ####### 1 ####### 2 =+ =ch 33. ms 2 or xt 1 =ch 167. ms 22 (ii) xt 2 1 2 =( ) 15 s ms 50 − 0 +af 50 ms()− 15 s or xt 2 =a 50 msf− 375 m (iii) For 40 s s≤≤ t 50 , x vt t 33 at 2 t 0 ####### 1 ####### 2 ####### = 40 50 40 ####### = F HG I KJ ####### + ()− + ()− area under vs from to 40 s s ma s sf or xt 3 375 1 250 2 t ####### 1 ####### 2 =+ + m m ej− −+ −5 0. m s 2 a 40 sf bg 50 m sa 40 sf which reduces to xtt 3 =− −bg 250 ms ej2 5.. m s 22 4 375 m (e) v === total displacement total elapsed time m s ms ####### 1875 ####### 50 ####### 37 5. P2 (a) Compare the position equation xt =+200 300.. − 400. t 2 to the general form xxvt afii =++ t ####### 1 ####### 2 2 to recognize that xi =200. m, vi =300. m s , and a =−800. ms 2. The velocity equation, vvafi =+ t , is then vtf = 300 ms−ch800.. ms 2. The particle changes direction when vf =0 , which occurs at t = 3 8 s. The position at this time is: x =+ F HG I KJ − F HG I KJ ####### 200 300 3 = ####### 8 ####### 400 3 ####### 8 ####### 256 2 .. m a m sf s ch. m s 2 s. m. (b) From xxvt afii =++ 1 t 2 2 , observe that when xx fi = , the time is given by t v a =− 2 i. Thus, when the particle returns to its initial position, the time is t = ####### − ####### − ####### = ####### 2300 ####### 800 ####### 3 ####### 4 ####### . ####### . ms ms 2 s af and the velocity is vf = − F HG I KJ ####### 300 800 3 = − ####### 4 .. ms ch ms 2 s 300. ms. Chapter 2 33 P2 In the simultaneous equations: vvat xx v vt xf xi x f i xi xf ####### =+ ####### − =+ R S | T| U V | W| ####### 1 ####### 2 ch we have vv vv xf xi xi xf ####### = − () ####### =+() R S | T | U V | W | ####### 560 420 ####### 62 4 ####### 1 ####### 2 ####### 420 ####### .. ####### .. ms s m s ch 2 ch ####### . So substituting for vxi gives 62 4 1 2 .. m m=+ () vvxf ch56 0 s 2 4 20. s s+() xf 4 20. ####### 14 9 ####### 1 ####### 2 .. ms=+ vxf ch 560 ms 2 ( ) 420. s. Thus vxf = 310. ms. P2 Take any two of the standard four equations, such as vvat xx v vt xf xi x f i xi xf ####### =+ ####### − =+ R S | T| U V | W| ####### 1 2 ch . Solve one for vxi , and substitute into the other: vvatxi = xf − x xxfi − = 1 v atvtxfx x − + f 2 ch. Thus xxvt atfi x − = f − x ####### 1 ####### 2 ####### 2. Back in problem 29, 62 4 4 20 1 2 .. .. m s m=( ) vxf −−ch 560 s s 2 () 420 2 vxf =62 4.. m m−49 4 = 310. 4 s ms. P2 (a) a vv t = fi − == 632 − =− 140 ####### 662 202 5280 ej 3600 ####### . ft s 22 m s (b) xvt atfi =+ = F HG I KJ ####### 1 −== ####### 2 ####### 632 ####### 5280 ####### 3600 ####### 140 1 ####### 2 2 a f a.. ff a662 1 40fa f 2649 t 198 m 34 Motion in One Dimension P2 (a) The time it takes the truck to reach 20 0. m s is found from vvafi =+ t. Solving for t yields t vv a = fi − =20 0 − 0 = 200 ####### 10 0 ####### . ####### . ####### . ms ms ms 2 s. The total time is thus 100 ++= 200 500 350 ..... s s s s (b) The average velocity is the total distance traveled divided by the total time taken. The distance traveled during the first 10 s is xvt 1 0200 2 ==F + 10 0 100 HG I KJ . ()=. m. With a being 0 for this interval, the distance traveled during the next 20 s is xvt a 2 i 1 t 2 2 =+ =( )( )+=20 0 20 0.. 0 400 m. The distance traveled in the last 5 s is xv 3 t 20 0 0 2 ==F + 500 500 HG I KJ . ()=.. m. The total distance xx x x =++= + += 123100 400 50 550 m , and the average velocity is given by v x t ####### == = 550 ####### 35 0 ####### 15 7 ####### . . ms. P2 We have vi =×200 10. m 4 s, vf =×600 10. m 6 s, xxfi − =×150 10. m− 2. (a) xxfi − =+ 1 vvti f 2 ch: t xx vv fi if ####### = ####### − ####### + ####### = ####### × ####### ×+× ####### =× − 2 2 1 50 10 − 2 00 10 6 00 10 ####### 498 10 2 46 ch ch. 9 ####### .. ####### . m ms ms s (b) vv axx 22 fi xfi =+ − 2 di: a vv x xx fi fi ####### = ####### − ####### − ####### = ####### ×−× ####### × ####### − =× 22 6 2 4 2 2 15 2 ####### 6 00 10 2 00 10 ####### 2 1 50 10 ####### 120 10 ####### () ####### .. ####### (. ) ####### . ms ms m ms ejej 2 36 Motion in One Dimension *P2 Let the glider enter the photogate with velocity vi and move with constant acceleration a. For its motion from entry to exit, xxvt at vt at v t vv at fixi x id d dd di d ####### =+ + ####### =+ + = ####### =+ ####### 1 ####### 2 ####### 0 1 ####### 2 ####### 1 ####### 2 2 ####### A ∆∆∆ 2 ####### ∆ (a) The speed halfway through the photogate in space is given by vv a vahs 22 i 2 i 2 vtd d 2 =+FHG IKJ=+ ####### A ####### ∆. vvahs =+ i 2 vtd ∆ d and this is not equal to vd unless a =0. (b) The speed halfway through the photogate in time is given by vva t ht =+ i d F HG I KJ ####### ∆ ####### 2 and this is equal to vd as determined above. P2 (a) Take initial and final points at top and bottom of the incline. If the ball starts from rest, vi =0, a =0500. ms 2 , xxfi − =900. m. Then vv axx v fi fi f ####### 22202 2 0 500 9 00 ####### 300 ####### =+ −=+ ####### = di ej. ####### .. ms m ms 2 (b) xxvt atfi i − =+ 1 2 2 ####### 900 0 ####### 1 ####### 2 ####### 0500 ####### 600 ####### .. 2 ####### . ####### =+ ####### = ms s 2 ej t t (c) Take initial and final points at the bottom of the planes and the top of the second plane, respectively: vi =300. ms, vf =0, xxfi − =15 00. m. vv axx 22 fi =+ 2 ch fi − gives a vv xx fi fi ####### = ####### − ####### − ####### = ####### − ####### () ####### =− 22 2 2 ####### 0300 ####### 2150 ####### 0300 ch a. f ####### . ####### . ms m ms 2. (d) Take the initial point at the bottom of the planes and the final point 8 m along the second: vi =300. ms, xxfi − =800. m, a =−0300. ms 2 vv axx v fi fi f ####### 222 3 00 2 2 0 300 8 00 4 20 ####### 205 ####### =+ −= +− = ####### = dibg.. f. ####### .. ms ms m m s ms 222 Chapter 2 37 P2 Take the original point to be when Sue notices the van. Choose the origin of the x -axis at Sue’s car. For her we have xis =0, vis =30 0. ms, as =−200. ms 2 so her position is given by xt x vtsi ()= + + = sissat t + − t ####### 1 ####### 2 ####### 30 0 ####### 1 ####### 2 22 a .. msf ch 200 2. ms For the van, xiv =155 m , viv =500. ms, av =0 and xt x vtvi ()= + + = + vivv 1 at t + 2 2155 a5 00. msf 0. To test for a collision, we look for an instant tc when both are at the same place: 30 0 155 5 00 0250155 2 2 ####### .. ####### .. tt t tt cc c cc ####### −= + ####### =− + From the quadratic formula tc = ####### ±( )− () ####### = ####### 25 0 25 0 4 155 ####### 2 ####### 13 6 ####### .. 2 . s or 11 4. s. The smaller value is the collision time. (The larger value tells when the van would pull ahead again if the vehicles could move through each other). The wreck happens at position 155 m m+(af5 00. s m11 4 )= 212. *P2 As in the algebraic solution to Example 2, we let t represent the time the trooper has been moving. We graph xt car=+ 45 45 and xt trooper= 15. 2. They intersect at t = 31 s. x (km) t (s) 10 20 30 40 ####### 0. ####### 1 ####### 1. car police officer ####### FIG. P2. Chapter 2 39 P2 We have ygfi =− 1 t ++ vtyi 2 2 0 =−−ch 490 ... ms 2 tt 2 af 800 ms+ 300 m. Solving for t , t = ±+ − ####### 800 640 588 ####### 980 ####### .. ####### . ####### . Using only the positive value for t , we find that t = 179. s. P2 (a) yyvt atfi i − =+ 1 2 2 : 400 ..=( ) 150 v −()()490 150.. 2 i and vi = 10 0. ms upward. (b) vvafi =+= t 10 0−()()=9 80 1 50 −4 68... m vf = 468. m s downward P2 The bill starts from rest vi =0 and falls with a downward acceleration of 980. ms 2 (due to gravity). Thus, in 0 s it will fall a distance of ∆ yvt gt = i − 1 = − ()=− 2 20 ch4 90... ms 2 0 20 s 2 0 20 m. This distance is about twice the distance between the center of the bill and its top edge a≅8 cmf. Thus, David will be unsuccessful. *P2 (a) From ∆ yvt at =+ i ####### 1 ####### 2 2 with v i =0 , we have t y a ####### == ####### ()− ####### − ####### = ####### 2223 ####### 980 ####### 217 a∆f m ms 2 s . ####### .. (b) The final velocity is vf =+ 09 ch−... 80 ms 2 ()= 17 s 2 − 12 ms 2. (c) The time take for the sound of the impact to reach the spectator is t y sound v sound m 340 m s == =×∆ 23 676 10.,− 2 s so the total elapsed time is t total=+× ≈2 17.. s s6 76 10− 2 2 23.. s 40 Motion in One Dimension P2 At any time t , the position of the ball released from rest is given by yh gt 1 1 2 2 = −. At time t , the position of the ball thrown vertically upward is described by yvt g 2 i 1 t 2 2 = −. The time at which the first ball has a position of y 1 h 2 = is found from the first equation as h hgt 2 ####### 1 ####### 2 = − 2 , which yields t h g =. To require that the second ball have a position of y 2 h 2 = at this time, use the second equation to obtain h v h g g h 2 i g ####### 1 ####### 2 ####### = − F HG I KJ . This gives the required initial upward velocity of the second ball as vgi = h. P2 (a) vvgfi = − t : vf =0 when t =300. s, g =980. ms 2. Therefore, vgti == ( )=ch9 80 ms 2 3 00 s 29 4... ms. (b) yyfi − =+ 1 vvtf i 2 ch yyfi −= 1 = 2 bg29 4 af3 00 44 1 ... ms s m *P2 (a) Consider the upward flight of the arrow. vv ayy y y yf yi y f i 22 2 ####### 2 ####### 0100 298 ####### 10 000 ####### 19 6 ####### 510 ####### =+ − ####### =+− ####### == di bg ms ej ms ms ms m 2 22 2 ####### . ####### . ####### ∆ ####### ∆ (b) Consider the whole flight of the arrow. yyvt at tt fiy =+ + i y =+ + − ####### 1 ####### 2 ####### 00 100 1 ####### 2 ####### 98 2 bg ms ej. m s 22 The root t =0 refers to the starting point. The time of flight is given by t == ####### 100 ####### 49 ####### 20 4 ms ms 2 s . ####### .. P2 Time to fall 3 m is found from Eq. 2 with vi =0, 300 1 2 .. m m= ch 980 s 2 t 2 , t =0782. s. (a) With the horse galloping at 10 0. m s , the horizontal distance is vt = 782. m. (b) t = 0782. s

Top 43 최신 대학 물리학 5 판 솔루션 The 194 Correct Answer

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최신대학물리학 5판 솔루션 북스힐, Raymond A Serway 업로드 XD – A Serwa.. Raymond A Serway최신대학물리학_5판 … – stewart solution – Stewart Solution

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최신대학물리학 5판 솔루션 북스힐, Raymond A Serway 업로드 XD

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