Conceptual Physics Paul Hewitt Solution Manual
Conceptual physics 12th edition hewitt solutions manual. 1. 1000 m 1 min 60 min Conceptual Physics 12th Edition Hewitt Solutions Manual Full clear download (no error formatting) at: hewitt-solutions-manual/ Conceptual Physics 12th Edition Hewitt Test Bank Full clear download (no error formatting) at: hewitt-test-bank/ Solutions 3-1. (a) Distance hiked = b + c km. (b) Displacement is a vector representing Paul’s change in position.
Drawing a diagram of Paul’s trip we can see that his displacement is b + (–c) km east = (b –c) km east. B km displacement –c km (c) Distance = 5 km + 2 km = 7 km; Displacement = (5 km – 2 km) east = 3 km east. (a) From v d t v x.
T (b) v x.We want the answer in m/s so we’ll need to convert 30 km to meters and 8 min t to seconds: 30.0 km 1000 m 30, 000 m; 8.0 min 60 s 480 s. Then v x 30,000 m 63 m.1 km 1 min t 480 s s Alternatively, we can do the conversions within the equation: v x 30.0 km 1 km 63 m. T In mi/h: 8.0 min 60 s s 30.0 km 1 mi 18.6 mi; 8.0 min 1 h 0.133 h. Then v x 18.6 mi 140 mi.1.61 km 60 min t 0.133 h h 1 mi Or, v x 30.0 km 60 min 1 mi 140 mi.
Or, v x 30.0 km 1.61 km 140 mi. T 8.0 min 1 h 1.61 km h t 8.0 min 1 h h There is usually more than one way to approach a problem and arrive at the correct answer! (a) From v d t. v L.
T (b) v L 24.0 m 40 m. T 0.60s s 3-4. (a) From v d t v x. T (b) v x 0.30 m 30 m. T 0.010 s s 3-5.
T t (b) v 2r 2(400m) 63 m. T 40s s. s s s m m 3-6. From v d t t h. V (b) t h 508 m 34 s. V 15 m (c) Yes.
At the beginning of the ride the elevator has to speed up from rest, and at the end of the ride the elevator has to slow down. These slower portions of the ride produce an average speed lower than the peak speed. Begin by getting consistent units. Convert 100.0 yards to meters using the conversion factor on the inside cover of your textbook: 0.3048 m = 1.00 ft. 1yard 1ftThen 100.0 yards 3ft 0.3048 m 91.4 m. From v d t d 91.4 m. (b) t d 91.4 m 15s.
T v v v 6.0 m 3-8. Fromv d t d L. T v c (b) t L 1.00 m 3.33 10-9 s 3.33 ns.
(This is 31 billionths of a second!) v 3.00 108 m 3 3-9. From v d t d vt. (b) First, we need a consistent set of units.
Since speed is in m/s let’s convert minutes to seconds: 5.0 min 60s 300 s. Then d vt 7.5 m 300s 2300 m.1min s 3-10.
(a) v v0 vf v. From v d t d vt vt. 2 (c) d vt 2.0 s (1.5s) 1.5 m.
From v d d vt v0 vf t 0 v t vt. T (b) d vt 12 s (8.0s) 2 2 2 48 m. From v d d vt v0 vf t 0 v t vt.
T 2 2 2. m (b) First get consistent units: 100.0 km/h should be expressed in m/s (since the time is in seconds).100.0 km 1 h 1000 m 27.8 m. Then, d vt 27.8 s (8.0 s) 110m. H 3600 s 1 km s 2 2. s m 2 2 2 3-13. (a) a v v2 v1. T t (b) v 40 km 15 km 25 km.
Since our time is in seconds we need to convert km to m:h h h h s m 25 km 1hr 1000 m 6.94 m. Then a v 6.94 s 0.35 m.h 3600 s 1 km s t 20s s2 Alternatively, we can express the speeds in m/s first and then do the calculation: 11.1 m 4.17 m 15 km 1hr 1000 m 4.17 m and 40 km 1hr 1000 m 11.1 m. Then a s s 0.35 m. Hr 3600 s 1 km s hr 3600 s 1 km s 20s s2 3-14. (a) a v v2 v1. T t (b) To make the speed units consistent with the time unit we’ll need v in m/s: v2 v1 4.17 m v v2 v1 20.0 km 5.0 km 15.0 km 1hr 1000 m 4.17 m. Then a 0.417.
H h h 3600 s 1 km s t 10.0 s s2 An alternative is to convert the speeds to m/s first: v1 5.0 km 1hr 1000 m 1.4 m; v2 20.0 km 1hr 1000 m 5.56 m.h 3600 s 1 km s h 3600 s 1 km s Then a v2 v1 5.56 s 1.4 s 0.42 m m m. T 10.0 s s2 m m (c) d vt v1 v2 t 1.4 s 5.56 s 10.0 s 35 m.
Or, 2 s 2 s2d v1t 1 at2 1.4 m (10.0 s) 1 0.42 m (10.0 s)2 35 m. (a) a v vf v0 t t v 26 m 0 v v. T t (b) a s 1.3 m. T 20s s2 m m (c) d? From v d d vt v0 vf t 26 s 0 s 20 s 260 m.
T 2 s 2 2 s2 Or, d v0t 1 at2 26 m (20 s) 1 1.3 m (20 s)2 260 m. Lonnie travels at a constant speed of 26 m/s before applying the brakes, so d vt 26 m (1.5 s) 39 m. (a) a v vf v0 t t v 72 m 0 v v. T t (b) a s 6.0 m.
T 12 s s2. s s 24 m m (c) d? From v d d vt v0 vf t 72 s 0 s (12 s) 430 m.
T 2 2 s 2 2 s2 Or, d v0t 1 at2 72 m (12 s) 1 6.0 m (12 s)2 430 m. From v d t d L 2L.
2 t v (b) t 2L 2(1.4 m) 0.19 s. Vf v0 v v 15.0 m 3-18. (a) v v0 vf v.
2 2 350 m (b) v s 175 m.Note that the length of the barrel isn’t needed—yet! 2 (c) From v d s t d L 0.40 m 0.0023 s 2.3 ms.
Conceptual Physics Hewitt Pdf
T v v 175 m 3-19. (a) From v d d vt v0 vf t = v0 v t. T 2 2 m m (b) d v0 v t = 25 s 11 s (7.8 s) 140 m. There’s a time t between frames of 1 s, so v= d x 24 1 x. (That’s 24x per24 second.) (b) v 24 1 x 24 1 (0.15 m) 3.6 m.
T 1 s s s s s 3-21. Since time is not a part of the problem we can use the formula vf 2 v0 2 2ad and v2 solve for accelerationa.
Then,withv0= 0 and d = x, a. 2x (b) a v 1.8 10 s 1.6 1015 m 2 7 m 2. 2x 2(0.10 m) s2 vf v0. 1.8 10 s 0 s (c) t?
From vf v0 at t a 7 m m 1.6 1015 m s2 1.1 10-8 s 11 ns. Or, from v d t d L 2L 2(0.10 m) 1.1 10-8 s.
2 1.8 10 s t v vf v0 (v 0) 7 m. m d v0 vf 2d 3-22. From v t 2 t with v0 0 vf t. From d v0t 1 at2 with v0 0 d 1 at2 a.2 2d 2(402 m) m 2 t2 2d 2(402 m) m (c) vf t 181; a 40.6. 4.45 s s t2 (4.45 s)2 s2 3-23. From v d d vt v0 vf t = v V t.
T 2 2 m m (b) d v V t = 110 s 250 s 2 2 (3.5 s) 630 m. Let's choose upward to be the positive direction. Vf v0 0 v v From vf v0 at with vf 0 and a g t (b) t v 32 s 3.3 s. A g g g 9.8 m s2 2 m 2 (c) d? From v d d vt v0 vf t v 0 v v 32 s 52 m. T 2 2 g 2g 2 9.8 m s2 2 s 2 s2We get the same result with d v0t 1 at2 32 m (3.3 s) 1 9.8 m (3.3 s)2 52m.
When the potato hits the ground y = 0. From d v0t 1 at2 y v0t 1 gt2 0 t v0 1 gt v0 1 gt.2 2 2 2 s 2 s 2 s 1000 m 2 1.61 km 1 h h (b) v0 1 gt 1 9.8 m (12 s) 59 m.In mi/h, 59 m 1 km 1 mi 3600 s 130 mi. Choose downward to be the positive direction. From From d v0t 1 at2 with v0 0, a g and d h h 1 gt2 t 2h. 2 2 g (b) t 2h 2(25m) 2.26 s 2.3s. s g 9.8 m s2 s2 s (c) vf vo at 0 gt 9.8 m (2.26 s) 22 m. S2 2 2 m m Or, from 2ad vf v0 with a g, d h, and v0 0 vf 2gh 2 9.8 (25 m) 22.
s m 0 0 0 0 0 0 3-27. Let’s call upward the positive direction.
Since the trajectory is symmetric, vf = –v0. Gt Then from vf v0 at, with a g v0 v0 gt 2v0 gt v0 2. 2gt 9.8 m (4.0s) m (b) v0 2 20. From v d d vt v0 vf t v0 t 20 s t 2 2 (2.0 s) 20 m. 2 We use t = 2.0 s because we are only considering the time to the highest point rather than the whole trip up and down. Let’s call upward the positive direction.
Since no time is given, use vf 2 v0 2 2ad with a = –g, vf = 0 at the top, and d = (y – 2 m). V2 2(g)(y 2m) v 2g(y 2m).
S2 s s (b) v0 2g(y 2 m) 2 9.8 m (20 m 2 m) 18.8 m 19 m. (a) Taking upward to be the positive direction, from 2 2 2 2ad vf v0 with a g and d h vf v0 2gh. So on the way up vf v2 2gh. (b) From above, on the way down vf v2 2gh, same magnitude but opposite direction as (a). 2 2gh v v0 v2 2gh (c) From a vf v0 t t vf v0 v0 0.
A g g m m 2 m 2 m s2 m vf v0 s a 9.5 s 16 s 9.8 m (d) vf v0 2gh 16 s 2 9.8 (8.5 m) 9.5. Taking upward to be the positive direction, from 2 2 2 2ad vf v0 with a g and d h vf v0 2gh.
The displacement d is negative because upward direction was taken to be positive, and the water balloon ends up below the initial position. The final velocity is negative because the water balloon is heading downward (in the negative direction) when it lands. 2 2gh v v0 v2 2gh (b) t =? From a vf v0 t t vf v0 v0 0. A g g.
(c) vf =? Still taking upward to be the positive direction, from 2 2 2 2 2 2ad vf v0 with initial velocity = –v0,a g and d h vf v0 2gh vf v0 2gh. We take the negative square root because the balloon is going downward. Note that the final velocity is the same whether the balloon is thrown straight up or straight down with initial speed v0. v0 v0 2 4 (h) 2 2 m 2 2 m 2 m m s2 (d) vf v0 2gh 5.0 s 2 9.8 (11.8 m) 16 s for the balloon whether it is tossed upward or downward. For the balloon tossed upward, m m t vf v0 16 s 5 s 2.1 s.
A 9.8 m s2 3-31. (a) Call downward the positive direction, origin at the top. From d v0t 1 at2 with a g, d ² y h h v0t 1 gt2 1 gt2 v0t h 0.2 From the general form of the quadratic formula x 2 b b2 4ac 2a 2 we identify g a g, b v, and c h, which gives t 2 v0 v0 2 2gh. 2 0 g g To get a positive value for the time we take the positive root, and get v0 + v0 2 + 2gh t. G (b) From 2 2 2 2 2 2ad vf v0 with initial velocity v0,a g and d h vf v0 2gh vf v0 2gh. V0 v0 2 2gh Or you could start with vf v0 at v0 g v0 g + 2gh.
M m 29.8 m (3.5 m) (c) t v0 v0 2 2gh g 3.2 s 3.2 s s2 9.8 m s2 0.58 s.; 2 m 2 m m s2 vf v0 2gh 3.2 s 2 9.8 (3.5 m) 8.9 s 3-32. (a) From d v0t 1 at2 a 2(d v0t). T2 (b) a 2(d v0t) 2120 m 13 s 5.0 s 4.4 m. T2 (5.0 s)2 s2 s s2 s (c) vf v0 at 13 m 4.4 m (5s) 35 m. (d) 35 m 1 km 1 mi 3600 s 78 mi. This is probably not a safe speed for driving ins 1000 m 1.61 km 1 h h an environment that would have a traffic light!
(a) From x vt v0 vf t v 2x v. 2 f t 0 x x (b) a vf v0 (2 t v0 )v0 2 t 2v0 2 x v0 t t t t2 t.
2 2 s v0 v0 2 4 (h) 2 1 0 2 1 v v a t 2 2 1 2x 2(95m) m m (c) vf v0 t 13 3.0. 11.9s s s m m m a 2 x v0 2 95m 13 s 0.84 m or a vf v0 3.0 s 13 s 0.84 m. T2 t (11.9s)2 11.9s s2 t 11.9s s 2 2 2 3-34. (a) From 2ad vf v0 with d L vf v0 2aL. This is Rita’s speed at the bottom of the hill. To get her time to cross the highway: From v d t (b) t d 25m 1.54s. V0 2aL v0 2aL 3.0 m 2 2 1.5 m (85m)s2 3-35.
(a) Since v0 is upward, call upward the positive direction and put the origin at the ground. Then From d v0t 1 at2 with a g, d ² y h h v0t 1 gt2 1 gt2 v0t h 0.2 From the general form of the quadratic formula x 2 b b2 4ac 2a 2 we identify g a g, b v, and c h, which gives t 2 v0 v0 2 2gh. 2 0 g g 2 2 2 2 2 (b) From 2ad vf v0 with a g and d h vf v0 2gh vf v0 2gh. M m 29.8 m (14.7m) (c) t v0 v0 2 2gh g 22 s 22 s s2 9.8 m s2 0.82s or 3.67s.
So Anthony has to have the ball leave his had either 0.82s or 3.67s before midnight. The first time corresponds to the rock hitting the bell on the rock’s way up, and the second time is for the rock hitting the bell on the way down.
2 m 2 m m s2 vf v0 2gh 22 s 2 9.8 (14.7m) 14 s. The rocket starts at rest and after time t1 it has velocity v1 and has risen to a height h1.
Taking upward to be the positive direction, from vf v0 at with v0 0 v1 at1. From d v t 1 at2 with h1 d and v0 0 h1 1 at2. For this stage of the problem the rocket has initial velocity v1, vf = 0, a = –g and the distance risen d = h2. 2 2 2 2 f 0 0 v2 2 (at1)2 2 2 1. 1 v 2 1 1 a 2 From 2ad vf v0 d h2 2a 2(g) 2g.
2g 2g (d) tadditional =? To get the additional rise time of the rocket: From a vf v0 t vf v0 0 v1 at1. T additional a g g (e) The maximum height of the rocket is the sum of the answers from (a) and (b) = 2 hmax h1 h2 1 at2 a2 t1 2g 1 at2 1 g. 2 (f) tfalling =? Keeping upward as the positive direction, now v0 = 0, a = –g and d = –hmax.
From d v0t 1 at2 hmax 1 (g)t2 2 2hmax 2 2 2 at1 1g at2 (g a) t tfalling 1 2 g g a 1 g2 at1 t1 a(g + a) 1 g (g) ttotal t1 tadditional tfalling t1 a(g + a). G g (h) vruns out of fuel v1 at1 120 m (1.70 s) 204 m; h 1 at1 120 (1.70 s) 173 m. S2 2 a2 t1 s 2 1 m 2 s 1 2 2 s2 2 120 m (1.70 s)2 hadditional h2 2g 2123 m. S2 2 9.8 m tadditional at1 g 120 m (1.70 s) s2 9.8 m s2 20.8 s. Hmax 173 m + 2123 m 2296 m 2300 m. Tfalling 2hmax g 2(2300 m) 9.8 m s2 21.7 s. Ttotal t1 tadditional tfalling 1.7 s 20.8 s 21.7 s 44.2 s.
V total distance x x 2x 1.14 x. Total time t 0.75t 1.75t t (b) v 1.14 x 1.14 140 km 80 km. T 2 hr hr 3-33. (a) v total distance. From v d d vt. Total time dwalk djog So v t vwalktwalk vjogtjog v(30 min) 2v(30 min) 3v(30 min) 1.5 v. Twalk tjog twalk tjog 30 min 30 min 2(30 min) (b) v 1.5v 1.51.0 m 1.5 m.s s (c) dto cabin vttotal v(twalk tjog ) 1.5 m (30 min +30 min) 60 s 5400 m = 5.4 km.
s 1 min 3-34. (a) v total distance. From v d d vt. Total time t So v dslow dfast vslowtslow vfasttfast v(1 h)4v(1 h) 5v(1 h) 2.5 v. Tslow tfast tslow tfast 1h 1h 2 h (b) v 2.5v 2.525 km 63 km.h h.
1 2 1 2 1 2 s m 3-35. (a) v total distance.
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From v d t d x. Total time t v v So v d1 d2 2x 2x 2 v v xv v v v t1 t2 x x 1 1 v2 v1 2 v1v2 v(1.5v) 1.5v2 2 2 1.2v.Note that the average velocity is biased toward v2 v1 1.5v v 2.5v the lower speed since you spend more time driving at the lower speed than the higher speed. (b) v 1.2v 1.228 km 34 km.h h 3-36. From VAtti dAtti d Vt.The time that Atti runs = the time that Judy t Atti walks, which is t x. So d V x V x. V Atti v v m (b) X V x 4.5 s (150m) 450 m. v 1.5 m 3-37.
T 1.5 s s 3-38. Call upward the positive direction. From vf 2 v0 2 2ad with d h, vf 0 and a g 2 v 2 2 2 m 2 h vf 0 v0 v0 14.7s 11 m. S2 2a 2(g) 2g 2 9.8 m 3-39. From v d d vt v0 vf t 0 27.5 s (8.0 s) 110 m.
t 2 2 3-40. Let's take down as the positive direction.
From d v0t 1 at2 with v0 0 and a g d 1 gt2 2 2 t 2d 2(16 m) 1.8s. G 9.8 m s2 m m 3-41.
A v t vf v0 t 12 s 0 s 3 s 4 m. S2.
m m 3-42. A v t vf v0 t 75 s 0 s 2.5 s 2 30 m. S2 2 2 s2 3-43. With v0 0, d v0t 1 at2 becomes d 1 at2 1 2.0 m (8.0 s)2 64 m. m m s 2d 3-44. With v0 0, d v0t 1 at2 becomes d 1 at2 a 2(5.0 m) 2.5 m. 2 2 2 2 2 t2 s2 (2.0 s)2 s2 3-45.
With v0 0, d v0t 1 at2 becomes d 1 at2 1 3.5 m (5.5 s)2 53 m. Here we’ll take upwards to be the positive direction, with a =g and vf = 0. S2 s From vf 2 v0 2 2ad v0 2 vf 2 2(g)d v0 2gd 2 9.8 m (3.0 m) 7.7 m. We can calculate the time for the ball to reach its maximum height (where the velocity will be zero) and multiply by two to get its total time in the air. Here we’ll take upward to be the positive direction, with a=g.
From a vf v0 t vf v0 v0 v0 18 s 1.84 s.This is the time to reach t a g g 9.8 m s2 the maximum height. The total trip will take 2 1.84 s = 3.7 s, which is less than 4 s. Alternatively, this can be done in one step with by recognizing that since the trajectory is symmetric vf = –v0. Then from vf v0 at, with a g v0 v0 gt 2v0 gt t 2v0 218 s 3.7 s. G 9.8 m s2 3-48. Since she throws and catches the ball at the same height, vf v0. Calling upward the positive direction, a = –g.
2gt 9.8 m (3.0 s) m From vf v0 at v0 v0 (g)t 2v0 gt v0 2 15. For a ball dropped with v0 = 0 and a = +g (taking downward to be the positive direction), fallen, 1st second 2 2 s2 d v0t 1 at2 1 9.8 m (1 s)2 4.9 m.
At the beginning of the 2nd second we have v0 = 9.8 m/s so fallen, 2nd second 2 s 2 s2 d d nd v0t 1 at2 9.8 m (1 s) 1 9.8 m (1 s)2 14.7 m. The. 2 2 ratio fallen, 2 second 14.7 m 3. More generally, the distance fallen from rest in a time dfallen, 1st second 4.9 m t is d 1 gt2. In the next time interval t the distance fallen is dfrom time t to 2t v0t 1 at2 (gt)t 1 gt2 3 gt2. The ratios of these two distances is2 2 2 3 2 dfrom time t to 2t 2 gt 3. Dfrom rest in time t 1 gt2.
s s d d 2 v 0 3-50. Call upward the positive direction. From vf 2 v0 2 2ad with d h, vf 0 and a g 2 v 2 2 2 m 2 h vf 0 v0 v0 1,000s 51,000 m 50 km.
S2 2a 2(g) 2g 2 9.8 m 3-51. With d h, v0 22 m, a –g and t 3.5 s, d v0t 1 at2 becomess s 2 2 s2h (22 m )(3.5 s) 1 9.8 m (3.5 s)2 17 m 3-52. From v d t d 65 m 5.0 s. T v 13 m m m 3-53. From a v t vf v0 t t vf v0 a 28 s 0 s 7.0 m s2 4.0 s. From v d t d d 2d.
2 t v vf v0 v (b) a =? With v0 0 and vf v, vf 2 – v0 2 2ad becomes a. 2d 2 m 2 (c) t 2d 2(140 m) 10 s; a v 28 s 2.8 m. V 28 m 2d 2(140 m) s2 3-55. From v d d vt v0 vf t 0 s 25 s m t 2 2 m (5.0 s) 63 m. 2462 mi 1 km 3-56. From v t 0.621 mi 8.5 min.
T v 28,000 km 1 h h 60 min 3-57. With vf 0, vf 2 – v0 2 2ad becomes mi 1 km 1000 m 1 h 2 v 2 a 2d 220 h 0.621 mi 1 km 3600 s 2(800 m) –6.05 m s2 –6 m.
From d v0t 1 at2 1 at2 v0t d 0. From the general form of the quadratic formula2 b b2 4ac x 2a.
2 2 we identify a 1 a, b v0, and c h, which gives v0 v0 2 41 a(d) 2 t 2 v0 v0 2ad. To get a positive answer for t we take a a 45 m 45 m 2 23.2 m (440m) s the positive root, which gives us t s s2 3.2 m s2 7.7s. g 2 3-59. The candy bar just clears the top of the balcony with height 4.2m + 1.1m = 5.3 m. With vf 0, vf 2 – v0 2 2ad with v0 and ² y positive and a g v0 2 vf 2 2(g)h s2 s s v0 2gh 2 9.8 m (5.3m) 10.19 m 10.2 m.
The total time is the time for the way to the top of the balcony rail plus the time to fall 1.1 m to the floor of the balcony. 2d 2(5.3 m) tup? From d vft 1 at2 with vf 0 and a g d 1 (g)t2 tup 1.04 s. 2 2 g 9.8 m s2 2h 2(1.1 m) tdown? From d v0t 1 at2 with v0 0, a g and d ² y h h 1 (g)t2 tdown 02 2 g 9.8 m s2 So ttotal tup tdown 1.040 s 0.47 s 1.51s. An alternative route is: Since v0 is upward, call upward the positive direction and put the origin at the ground. Then From d v0t 1 at2 with a g, d ² y 4.2m d v0t 1 gt2 1 gt2 v0t d 0.2 From the general form of the quadratic formula x 2 b b2 4ac 2a 2 we identify v0 v0 2 4 (d) 2 2gd a g, b v, and c d, which gives t 2 v0 v0 2 10.19 m 0 g g 10.19 m 2 29.8 m (4.2m) s s s2 9.8 m s2 0.57s or 1.51s.
The first answer corresponds to the candy reaching 4.2 m but not having gone over the top balcony rail yet. The second answer is the one we want, where the candy has topped the rail and arrives 4.2 m above the ground. Consider the subway trip as having three parts—a speeding up part, a constant speed part, and a slowing down part. Dtotal dspeeding up dconstant speed dslowing down. S2 0 2 2 s 2 For dspeeding up, v0 0, a 1.5 m and t 12 s, so d v t 1 at2 1 1.5 m (12 s)2 108 m.
For dconstant speed vt. From the speeding up part we had v0 0, a 1.5 m s and t 12 s s 2 s sso v v0 at 1.5 m (12 s) 18 m and so d 18 m (38 s) 684 m For dslowing down, vf 0, a –1.5 m and t 12 s, so d v t 1 at2 1 -1.5 m (12 s)2 108 m.
S2 f 2 2 s2. So dtotal dspeeding up dconstant speed dslowing down 108 m 684 m 108 m 900 m. One way to approach this is to use Phil’s average speed to find how far he has run during the time it takes for Mala to finish the race. From v d d v t 100.0 m (12.8 s) 94.1 m.
Since Phil has only t Phil Phil Mala 13.6 s traveled 94.1 m when Mala crosses the finish line, he is behind by 100 m 94.1 m 5.9 m 6 m. 600 h 3-60. The time for Terrence to land from his maximum height is the same as the time it takes for him to rise to his maximum height.
Let’s consider the time for him to land from a height of 0.6 m. Taking down as the positive direction: From d v0t 1 at2 with v0 0 and a g d 1 gt2 2 2 t 2d 2(0.6 m) 0.35s. G 9.8 m s2 His total time in the air would be twice this amount, 0.7 s. V d 1 mi 80 mi.
T 45 s 3 1 h s 3-62. V total distance. If we call the distance she drives d, then from v d t d. Total time t v So v dthere dback 2d 2d 2 2 vtherevback tthere tback d v d dv 1 v 1 v v back there vthere back there back vback vthere there back v v 40 km 60 km 2400km 2 2 h h 2 h 48 km. Note that the average velocity is biased 60 km 40 km 100 km h h h h toward the lower speed since Norma spends more time driving at the lower speed than at the higher speed. Conceptual Physics 12th Edition Hewitt Solutions Manual Full clear download (no error formatting) at: hewitt-solutions-manual/ Conceptual Physics 12th Edition Hewitt Test Bank Full clear download (no error formatting) at: hewitt-test-bank/ conceptual physics 12th edition answers conceptual physics 12th edition pdf conceptual physics 12th edition paul g.
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