Chapter 15 HW Solution
15.3
F=
ke ( 2e)( 79e) r2
19 60 N m 2 ( 158) ( 1. 10 C = 8. 109 99 2 C2 0 ( 2. 1014 m )
)
2
= 91 N
si ( repul on )
15.8
The magnitude of the repulsive force between electrons must equal the weight of an electron, Thus, kee2 r2 = m e
Formulas Given On the Tests: d = v o t + at2 V2 =V o 2 + 2ad F = ma a=v2/r V = V o + at F = kqQ/r2 F = qE I = Q/t W=Energy V = kQ/r V = PE/q V=Ed I = V/R P = IV Energy = 1/2CV2 R= L/A C = Q/V C = A/d F = q v B sin F = I l B sin B = I/(2r) B = nI -t/(RC) -
Problem Solutions
24.1
ybright = ym +1 ym = = 8 ( 632. 10
L
d m
( m + 1)
L
d
m=
L
d
2
9
0. 10 m 200
3
00 ) ( 5. m ) = 1. 10 58
m = 1. cm 58
24.2
(a) For a bright fringe of order m, the path difference is = m , where m = 0,1,2, At the location of the thir
Chapter 23 Solution
23.5 Since the mirror is convex, R < 0 . Thus, R = 0. m . With a real object, p > 0 , 550 so p = +10. m . The mirror equation then gives the image distance as 0
121 2 1 , or q = 0. m 268 = q R p 0. m 10. m 550 0
r ual and located 0. m
Chapter 22 Solution
22.9 (a) From Snells law, n2 =
00 n 0 n1 si 1 ( 1. ) si 30. n 52 = = 1. si 2 n si n19. 24
(b) 2 =
0
n2 c
=
632. nm 8 = 417 nm 1. 52
(c)
f=
0
=
3. 108 m s 00 = 4. 1014 H z in air and in syrup 74 632. 109 m 8
(d) v2 =
00 c 3. 108 m s = =
Chapter 21 Solution
21.43 From v = f , the wavelength is
00 v 3. 108 m s = = 00 = 4. 106 m = 4 000 km f 75 H z
The required length of the antenna is then,
L = 4 = 1000 km , or about 621 miles. Not very practical at all.
21.45
(a) The frequency of an elect
Problem Solutions
20.1 The magnetic flux through the area enclosed by the loop is
B = BA cos = B ( r2 ) cos0 = ( 0. T ) ( 0. m 30 25
2 9 ) = 5. 102 T m 2
20.2
The magnetic flux through the loop is given by B = BA cos where B is the magnitude of the magne
Problem Solutions
19.2 (a) For a positively charged particle, the direction of the force is that predicted by the right hand rule. These are: (a) (c)
i pl n ane ofpage and t l t o ef outoft page he
(b) (d)
i o t page nt he
i pl n ane ofpage and t ar t t o
Problem Solutions
18.2 (a)
R eq = R1 + R 2 + R 3 = +4. + 8. + 12 = 24 0 0
(b) The same current exists in all resistors in a series combination.
I=
V 24 V 0 = = 1. A R eq 24
(c) If the three resistors were connected in parallel,
1 1 1 1 1 1 1 + + = + + R
Physics 102 Chapter 17 Homework Solution
17.1 The charge that moves past the cross section is Q = I( t) , and the number of electrons is
n=
Q I( t) = e e
3
C s) ( 10. m i ) ( 60. s m i ) 0n 0 n = 3. 1020 el r 00 ect ons 19 1. 10 C 60 The negatively charge
Chapter 16
16.3
Problem Solutions
The work done by the agent moving the charge out of the cell is
W input = W
fe d il
= ( PEe ) = + q( V )
J = ( 1. 1019 C ) + 90 103 = 1. 1020 J 60 4 C
16.4 (extra)
q=
PEe = q( V ) = q V f V i , so
PEe 1. 10 J 92 = = 3. 1