Page 1

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Chapter 2 2.1

F E V w 4. a = 5. R = 6. w = g m I 1h 2 PE XL PE v 7. g = 8. h = 9. h = 10. f = 11. w = Pt 12. F = pA mh mg 2g 2p L v f - vi W F 2( KE ) 2( KE) W 2 13. t = 14. A = 15. m = 16. v = 17. s = 18. a = 2 P P m F t v V -E E -V v 2 - v1 p Rd 2 19. I = or I = 20. t = 21. P = 22. L = -r r a 2R k 5F -160 5( F - 32) 9 1 23. C = or C = 24. F = C + 32 25. f = 9 9 5 2p CX C Q1 + Q1P Q1 RA 26. L = 27. R3 = RT - R1 - R2 - R4 28. Q2 = or Q2 = + Q1 r P P 2 2 2 2 v - vi v - vi I N VN 29. I P = s s 30. N S = s P 31. v i = 2v avg - v f 32. a = or 2( s - si ) 2 s - 2si NP VP 2 2 Ft mV2 - Ft QJ v - v i + 2asi 33. s = 34. V1 = V2 35. R = 2 or V1 = 2a m m I t A V kL 3V 1 2 36. x i = x - v it - at 37. r = 38. r = 39. d = 40. r = p ph R ph 2 QJ Fr 41. I = + 42. I = + Rt m 1. s = vt

2. v = at

3. m =

(b) 162 cm 2

2. (a) V= lwh

(b) 105 in.

5. (a) c = P – a – b

2.2 1. (a) A= bh 4. (a) b =

P 4

(b) 158 ft

7. (a) r =

C 2p

P - 2a (b) b = 2 V 11. (a) h = 2 (b) 6.11 m pr A 14. (a) r = (b) 12.15 m p 9. (a) b =

P -a 2

20. (a) h =

(b) 33.2 km

12. (a) h =

2A b

4 3 pr 3

A h

(b) 7.50 cm

6. (a) d =

C p

(b) 26.0 m

10. (a) V = pr 2 h

(b) 1,460,000 m 3

V (b) 154 m 2 h 3V (b) 21.6 in. 16. (a) r = (b) 13.2 m ph 3V (b) 70,690 m 3 19. (a) B = h (b) 5.80 cm

15. (a) b = A 18. (a) V =

2A a+b

A 2p r

3. (a) b =

(b) 6.0 cm

8. (a) h =

(b) 10.9 yd

17. (a) C = 2 pr (b) 121.6 m (b) 122.4 ft 2

(b) 4420m 3

13. (a) B =

(b) 11.40 m

2.3 1. V = 1wh = ( 36.0cm )( 30.0cm )( 24.0cm ) = 25, 900cm 2. V = pr 2 h = p ( 2.10in.) ( 7.50in.) = 104in 3 3. V = 2

æ 11.40cm ö 4. V = pr 2 h = p ç ÷ è 2 ø

2

( 24.00cm) = 2450cm 3

3

1 2 1 2 pr h = p ( 5.40cm) ( 9.30cm ) = 284cm 3 3 3 2

é 11.40cm ù 2 5. A = pr 2 = p ê ú = 102.1cm 2 ë û

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6. A = pdh = p (11.40cm )( 24.00cm ) = 859.5cm

2

æ 1 ö é 1 ù 3 7. V = ç1w + bh÷ h' = ê(22.0 ft )(10.0 ft ) + (22.0 ft )( 4.70 ft ) ú(37.0 ft ) = 10,100 ft è ø 2 2 ë û éa +b ù é 3.70 ft + 6.80 ft ù 2 h=ê 8. A = ê ú ú(19.3 ft ) = 101 ft 2 ë 2 û ë û 9. V = 1wh = ( 9.00 ft )(12.0 ft ) (8.00 ft ) = 864 ft 11. A =

2

é 3.25cm ù 2 10. A = pr 2 = p ê ú = 8.30cm ë 2 û

3

1 1 2 bh = ( 4.00cm )( 6.00cm ) = 12.0cm 2 2

12. c = a 2 + b 2 =

( 4.00cm ) 2 + ( 6.00cm) 2 = 7.21cm

éæ 3.50cm ö 2 æ 3.20cm ö 2 ù 13. A = p r12 - r2 2 = p êç ÷ -ç ÷ ú = 1.58cm 2 êëè 2 ø è 2 ø úû

(

)

14. V =

4p r 3 4 3 = p (8.00m) = 2140m 3 3 3

15. w =

16. h =

V 192m = = 6.00 m 1w ( 8.00m )( 4.00m )

17. V = p r h = p (2.00cm) (4.20cm) = 52.8cm 2

18. V = p r h = p (3.90cm) (8.00cm) = 382cm 2

2

19. d =

C 29.5m = = 9.39 m p p

20. h =

V

pr 2

=

100 0m 3 æ 9.39m ö pç è 2 ÷ø

2

A 900m 2 = 36.0m 1 25.0m 2

3

3

= 14.4m

æ 1m ö ÷ = 137m 21. Distance=C (no.of rev) = pd (no.of rev ) = p ( 30.0cm )(145) ´ ç è 100cm ø æ 1L ö ÷ = 85L 22. A = Ch = (29.5m )(14.4m )ç è 5.0m 2 ø

24. r =

V = ph

1 ft 3 7.50gal = 22.5 ft p ( 42.0 ft )

500, 00gal ´

26. Atotal = Arec tan lg e - Atrapezoid 27. V = Vcylinder + Vcone 28. r =

A = p

23. h =

V

pr 2

50 0,000 gal ´ =

1 ft 3 7.50gal

p (18.0 ft )

= 65.5 ft

2

A1 (12.0 ft ) (15.0 ft ) 180 ft = = = 60 panels A2 (1.00 ft )( 3.00 ft ) 3.00 ft 2 2

25.

æ 1.90cm + 2.20cm ö A = (3.50cm) (2.00cm) - ç ÷ ( 0.400cm ) = 6.18cm 2 è ø 2

1 2 2 3 V = p (1.50m ) ( 2.75) + p (1.50m ) ( 2.00m ) = 24.1m 3

3.05m 2 = 0.985m;yes p

29. V = 1wh = (12.0 ft )( 20.0 ft ) (0.500 ft ) ´

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1yd

3

27 ft

3

= 4.44 yd 3


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3

30. 1 =

3

V 2.00 yd 27ft = ´ = 40.5 ft wh (4.00 ft )( 0.333 ft ) 1yd 3

31. V = 1wh - pr 2 h = (8.00in .) (8.00in.)( 6.00in.) - p ( 2.50in.) ( 6.00in.) = 266in3 2

32. V = pr 2 h - p r 2 h = p ( 25.0cm ) ( 60.0cm ) - p (10.0cm ) ( 60.0cm ) = 99, 000cm 3 2

2

Chapter 2 Review Questions 1. c 2. b 3. a 4. (1) To find the volume of liquid storage tanks. (2) To determine the amount of concrete needed for a driveway. 5. As a shorthand way to designate different measured quantities of the same type. 6. Most mistakes are made in problem solving by missing needed information or misinterpreting the information given. 7. Making a sketch helps visualize what is happening in the problem. 8. The basic question. 9. The working equation is found by solving the basic equation for the unknown quantity. 10. Carrying the units through a problem shows whether the answer is the kind expected. 11. Making an estimate of the correct answer shows whether the solution is reasonable. Chapter 2 Review Problems 1. (a) m =

F

(b) a =

F

2. h =

v

2

3. v f =

a m 2g 5. b = P – a – c = 36 ft – 12 ft – 6 ft = 18 ft 6. a = 2

2s t

- vi

4. v =

2KE m

2 æ A - b ö = 2 æ 210m - 16.0m ö = 12.0m 7. r = A = 2.19m çè 15.0m è h 2ø 2 ÷ø p

9. h =

8. A = ½ b h = ½ (12.2cm) (20.0cm) = 122 cm2

3V

pr

2

3

3(314cm )

=

p (5.00cm)

2

= 12.0cm

10. c = a + b = (41.2mm) + (9.80mm) = 42.3mm 11. A = 2p rh = 2p (7.20cm)(13.4cm) = 606cm 2

2

2

40.0cm = 2(14.0cm) + 2w 12.

w = 6.0cm

2

13. r =

V

ph

2100m

=

3

p 17.0m

= 6.27 m

2

2A

2(88.6m )

= = 14.4 m b 12.3m 15. V = V1 - V2 = (9.0cm × 6.0cm × 12cm) - (6.0cm × 3.0cm × 12cm) = 430cm 14. h =

16. A = A1 - A2 = (40.0cm × 120cm) - (10.0cm × 12.0cm) = 4680m

2

Chapter 2 Applied Concepts 1. A property - Ahouse = l prop w prop - lhouse w house = (100 ft ) (200 ft ) - ( 35.0 ft )( 80.0 ft ) = 17,200 ft 2

( (

) )

2 $50.00 9 ft = $0.026 / yd 2 = 2.62 Ë / yd 2 17200 ft 1yd 2

2. V = h ´ w ´ l = (8.00 ft )(10.0 ft )( 32.0 ft ) = 2560 ft ; 3

3. VSolidBeam = lwh = ( 240 ft ) (8.00in.)(8.00in.) = 15400in 3

2560 ft (1 min ) = 2.13 ft 3 / s 20.0 min ( 60sec) 3

VIBeam = Vtop + Vvertical + Vbottom = (1.00in.)(8.00in.) (24 0in.) + (6.00in.) (1.00in )( 240in.) + (1.00in)( 8.00in.)( 240in.) = 5280in3 3

Vsolid 15400in = 3 = 2.91 VIBeam 5280in

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4. # of balls wide=

w 16.8in. = = 2.10balls 2r 2´ 4.00in.

# of balls high=

h 16.8in. = = 4.20balls 2r 2´ 4.00in.

2 balls x 2 balls x 4 balls = 16 balls # of balls long= 5.

l 33.6in. = = 4.20balls 2´ r 2 ´ 4.00in.

(a) Vcylinder = p r h = p (1.53m ) ( 0.915m) = 6.73m 2

2

(

3

)

(b) m = DV = æè 7750 kg 3 öø 6.73m 3 = 522,000kg m

F = m ´ g = (522, 000kg )æè 9.80 m

ö = 512, 000N s 2ø

Full file at https://testbankuniv.eu/Applied-Physics-11th-Edition-Ewen-Solutions-Manual

Applied Physics 11th Edition Ewen Solutions Manual  

Full file at https://testbankuniv.eu/Applied-Physics-11th-Edition-Ewen-Solutions-Manual

Applied Physics 11th Edition Ewen Solutions Manual  

Full file at https://testbankuniv.eu/Applied-Physics-11th-Edition-Ewen-Solutions-Manual

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