Solution manual for applied physics 11th edition by ewen schurter gundersen

Solution Manual for Applied Physics 11th Edition by Dale Ewen, Neill Schurter, P Erik Gundersen Link full download solution manual: https://findtestbanks.com/download/s olution-manual-for-applied-physics11th-edition-by-ewen-schurtergundersen/ Link full download test bank: https://findtestbanks.com/download/te st-bank-for-applied-physics-11thedition-by-ewen-schurter-gundersen/ Chapter 2.1 s = vt PE g  13 t  PE h  mh W 23 C  26 L  r 5F 160 F r or C  20 t  T 36 x  x  v t  i i  37 r  at 21 P  24 F  1h 11 w = Pt A 22 L  2R C  32 k (b) 162 cm P v2  vi 32 a  2s  s i mV2  Ft or V1  38 r  V h (a) b  (a) r  35 R  39 d  I t 40 r  3V h kL R 11 (a) h  2V A (b) 7.50 cm (b) 33.2 km A 12 (a) h  (b) 12.15 m  17 (a) C = 2A (a) h  b (b) 10.9 yd (a) d  (b) 6.0 cm 10 (a) V  r2 h (b) 5.80 cm 18 (a) V  r (b) 121.6 m 20 (a) h  13 (a) B  V 16 (a) r  (b) 21.6 in r3 (b) 70,690 m 3 2A  (b) 1,460,000 m h 15 (a) b  A C (b) 26.0 m 2r A 14 (a) r  C (b) 6.11 m r2 (b) 122.4 ft (a) b  (b) 4420m (a) c = P – a – b 2 P (b) b   a P  2a v  vi2 s  2si or QJ m (a) V= lwh (b) 105 in (b) 158 ft (b) 154 m2 3V (b) 13.2 m h 3V 19 (a) B  h (b) 11.40 m a b 2.3 V  1wh  36.0cm30.0cm24.0cm  25, 900cm r2h   5.40cm 9.30cm  284cm3 3 V  r2h   2.10in. 7.50in.  104in 3 V  2  11.40cm  V  r h     t 2CX Q1 C Q 28 Q  Q1  Q1P or Q  2 h P v f  vi 25 f  2.2 (a) b  18 a  Fr m 42 I   (a) A= bh 12 F = pA  P Ft  QJ 41 I   Rt 10 f  m V w  I 31 vi  2vavg  v f 34 V1  V2  2a E R  XL a  R  R R I sNs VsN P 29 I P  30 NS  NP VP v  vi2  2asi 33 s  F m 2g 5F  32 a   27 R  R  w g v2 2L W 2KE 2KE 17 s  16 v  15 m  m F v  Rd v2  v1 P EV or I  RA h  mg 14 A  P VE 19 I  m  v = at    24.00cm  2450cm  A  r  11.40cm      2   102.1cm  A  dh   11.40cm24.00cm  859.5cm2    3 V  1w  bh h'  22.0 ft 10.0 ft   22.0 ft  4.70 ft  37.0 ft   10,100 ft     A  ab  3.70 ft  6.80 ft  h 19.3 ft  101 ft           3.25cm 2 V  1wh  9.00 ft12.0 ft8.00 ft  864 ft 11 A  bh  10 A  r2       8.30cm2  4.00cm6.00cm  12.0cm 2 12 c  a  b  4.00cm  6.00cm  7.21cm  3.50cm   3.20cm   13 A   r  r       1.58cm        2      4r   8.00 14 V  m 16 h  V  192m 900m2 15 w   6.00 m 17 V   r h   (2.00cm) (4.20cm)  52.8cm  25.0m 36.0m 8.00m 4.00m  1w A  2140m3 18 V   r h   (3.90cm) (8.00cm)  382cm 19 d  20 h  C   29.5m  9.39 m   V 100 0m3  14.4m  9.39m      r2    1m    137m 21 Distance=C no.of rev  dno.of rev   30.0cm145  100cm     ft   22 A  Ch  29.5m V 24 r   h total  ft  7.50gal  22.5 ft 42.0 ft A rec tan lg e 27 V  Vcylinder  Vcone trapezoid 50 0,000 gal  V 14.4m    85L  5.0m  500, 00gal  A 26 A  1L   23 h  r  7.50gal 18.0 ft   65.5 ft  A1  12.0 ft15.0 ft 180 ft  25   panels A2 1.00 ft3.00 ft 3.00 ft2  1.90cm  2.20cm  0.400cm  6.18cm A  3.50cm2.00cm        2 V  1.50m 2.75   1.50m 2.00m  24.1m3 3 yd A 28 r    3.05m  0.985m; yes  29 V  1wh  12.0 ft20.0 ft0.500 ft  27 ft 3 4.44 yd 30  V wh  2.00 yd 4.00 ft0.333 ft  27 ft yd  40.5 ft 31 V  1wh  r2h  8.00in 8.00in.6.00in.  2.50in. 6.00in.  266in3 32 V  r2h  r 2h   25.0cm 60.0cm  10.0cm 60.0cm  99, 000cm3 2 Chapter Review Questions c b a (1) To find the volume of liquid storage tanks (2) To determine the amount of concrete needed for a driveway As a shorthand way to designate different measured quantities of the same type Most mistakes are made in problem solving by missing needed information or misinterpreting the information given Making a sketch helps visualize what is happening in the problem The basic question 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 Review Problems F F v h  (a) m  (b) a  2g a m b = P – a – c = 36 ft – 12 ft – ft = 18 ft  A b   210m 16.0 m   a  h   2 2  15.0 m v f  2s t vi v  2KE m A     12.0 m r    2.19 m h  A = ½ b h = ½ (12.2cm) (20.0cm) = 122 cm 3V r  3(314 cm )  (5.00cm)  12.0cm 2 10 c  a  b  (41.2 mm)2  (9.80 mm)2  42.3mm 11 A  2rh  2 (7.20cm)(13.4 cm)  606cm 2 40.0cm  2(14.0cm)  2 12 13 r    6.0cm 2A  2(88.6 m )   14.4 m b 12.3m 14 h  V   2100m 17.0 m  6.27 m h 15 V  V  V  (9.0cm  6.0cm  12cm)  (6.0cm  3.0cm  12cm)  430cm 16 A  A 1 A 2 (40.0cm  120cm)  (10.0cm  12.0cm)  4680 m2 Chapter Applied Concepts Aproperty  Ahouse  l prop w prop  lhousewhouse  100 ft20 ft 35.0 ft80.0 ft  17,200 ft 9 ft  $0.026 / yd 17200 ft 1 yd  $50.00 2  2.62  / yd2 V  h  w  l  8.00 ft10.0 ft32.0 ft  2560 ft 3; VSolidBeam  lwh  240 ft8.00in.8.00in.  15400in   2560 ft 1 min  2.13 ft / s 20.0 60 sec  VIBeam  Vtop  Vvertical  Vbottom  1.00in.8.00in.24 0in. 6.00in.1.00in240in. 1.00in8.00in.240in.  5280in3 15400in Vsolid   2.91 VIBeam 5280in w 16.8in   2.10balls 2r  4.00in h 16.8in # of balls high=   4.20balls 2r  4.00in # of balls wide= balls x balls x balls = 16 balls l # of balls long=  33.6in 2 r  4.20balls  4.00in (a) Vcylinder   r 2h  1.53m 0.915m  6.73m    kg  (b) m  DV  7750 6.73m  522,000kg m 3  F  m  g  522, 000kg 9.80 m   512, 000N s 2