1. Conservation of momentum requires that the gamma ray particles move in opposite
directions with momenta of the same magnitude. Since the magnitude p of the
momentum of a gamma ray particle is related to its energy by p = E/c, the particles have
the sam
1. (a) The charge that passes through any cross section is the product of the current and
time. Since 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 103 C.
(b) The number of electrons N is given by q = Ne, where e is the magnitude of
1. Charge flows until the potential difference across the capacitor is the same as the
potential difference across the battery. The charge on the capacitor is then q = CV, and
this is the same as the total charge that has passed through the battery. Thus,
1. The vector area A and the electric field E are shown on the diagram below. The angle
between them is 180 35 = 145, so the electric flux through the area is
(
)
= E A = EA cos = (1800 N C ) 3.2 103 m cos145 = 1.5 102 N m 2 C.
2
b
g
2
2. We use = E A ,
1. (a) With a understood to mean the magnitude of acceleration, Newtons second and
third laws lead to
c6.3 10 kghc7.0 m s h = 4.9 10
m=
7
m2 a2 = m1a1
2
2
2
9.0 m s
7
kg.
(b) The magnitude of the (only) force on particle 1 is
2
qq
q
F = m1a1 = k 1 2 2 = 8
1. (a) Since the gas is ideal, its pressure p is given in terms of the number of moles n, the
volume V, and the temperature T by p = nRT/V. The work done by the gas during the
isothermal expansion is
W=
V2
V1
p dV = n RT
V2
V1
dV
V
= n RT ln 2 .
V
V1
We s
1. (a) The motion from maximum displacement to zero is one-fourth of a cycle so 0.170 s
is one-fourth of a period. The period is T = 4(0.170 s) = 0.680 s.
(b) The frequency is the reciprocal of the period:
f=
1
1
=
= 1.47 Hz.
T 0.680 s
(c) A sinusoidal wa
1. The air inside pushes outward with a force given by piA, where pi is the pressure inside
the room and A is the area of the window. Similarly, the air on the outside pushes inward
with a force given by poA, where po is the pressure outside. The magnitud
1. (a) The center of mass is given by
xcom = [0 + 0 + 0 + (m)(2.00) + (m)(2.00) + (m)(2.00)]/6.00m = 1.00 m.
(b) Similarly, ycom = [0 + (m)(2.00) + (m)(4.00) + (m)(4.00) + (m)(2.00) + 0]/6m = 2.00 m.
(c) Using Eq. 12-14 and noting that the gravitational e
1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a
distance r from the wire, is given by
B=
0i
2 r
.
With r = 20 ft = 6.10 m, we have
c4 10
B=
hb
2 b6.10 mg
7
T m A 100 A
g = 3.3 10
6
T = 3.3 T.
(b) This is about one-
1. We note that the symbol q2 is used in the problem statement to mean the absolute value
of the negative charge which resides on the larger shell. The following sketch is for
q1 = q2 .
The following two sketches are for the cases q1 > q2 (left figure) an
1. (a) Eq. 28-3 leads to
6.50 1017 N
FB
v=
=
= 4.00 105 m s .
19
3
eB sin
160 10 C 2.60 10 T sin 23.0
.
c
hc
h
(b) The kinetic energy of the proton is
K=
2
121
mv = 167 1027 kg 4.00 105 m s = 134 1016 J.
.
.
2
2
c
hc
This is (1.34 10 16 J) / (1.60 10 19
1. Our calculation is similar to that shown in Sample Problem 42-1. We set
K = 5.30 MeV=U = (1/ 4 0 )( q qCu / rmin ) and solve for the closest separation, rmin:
rmin
19
9
q qCu
kq qCu ( 2e )( 29 ) (1.60 10 C )( 8.99 10 V m/C )
=
=
=
4 0 K 4 0 K
5.30 106
1. If R is the fission rate, then the power output is P = RQ, where Q is the energy released
in each fission event. Hence,
R = P/Q = (1.0 W)/(200 106 eV)(1.60 10 19 J/eV) = 3.1 1010 fissions/s.
2. We note that the sum of superscripts (mass numbers A) must
1. According to Eq. 39-4 En L 2. As a consequence, the new energy level E'n satisfies
2
FG IJ = FG L IJ
H K H L K
En
L
=
En
L
2
=
1
,
2
which gives L = 2 L. Thus, the ratio is L / L = 2 = 1.41.
2. (a) The ground-state energy is
( 6.63 10 J s )
h2
E1 =
n2
1. From the time dilation equation t = t0 (where t0 is the proper time interval,
= 1 / 1 2 , and = v/c), we obtain
FG t IJ .
H t K
2
= 1
0
The proper time interval is measured by a clock at rest relative to the muon. Specifically,
t0 = 2.2000 s. We are
1. (a) For a given value of the principal quantum number n, the orbital quantum number
ranges from 0 to n 1. For n = 3, there are three possible values: 0, 1, and 2.
(b) For a given value of , the magnetic quantum number m ranges from to + . For
= 1 , the
1. The number of atoms per unit volume is given by n = d / M , where d is the mass
density of copper and M is the mass of a single copper atom. Since each atom contributes
one conduction electron, n is also the number of conduction electrons per unit volu
1. (a) Let E = 1240 eVnm/min = 0.6 eV to get = 2.1 103 nm = 2.1 m.
(b) It is in the infrared region.
2. The energy of a photon is given by E = hf, where h is the Planck constant and f is the
frequency. The wavelength is related to the frequency by f = c,
1. (a) The flux through the top is +(0.30 T)r2 where r = 0.020 m. The flux through the
bottom is +0.70 mWb as given in the problem statement. Since the net flux must be zero
then the flux through the sides must be negative and exactly cancel the total of
1. With speed v = 11200 m/s, we find
K=
121
mv = (2.9 105 ) (11200) 2 = 18 1013 J.
.
2
2
2. (a) The change in kinetic energy for the meteorite would be
(
)(
1
1
K = K f K i = K i = mi vi2 = 4 106 kg 15 103 m/s
2
2
)
2
= 5 1014 J ,
or | K |= 5 1014 J . The
1. The x and the y components of a vector a lying on the xy plane are given by
ax = a c os ,
a y = a sin
where a =| a | is the magnitude and is the angle between a and the positive x axis.
(a) The x component of a is given by ax = 7.3 cos 250 = 2.5 m.
(b
1. Using the given conversion factors, we find
(a) the distance d in rods to be
d = 4.0 furlongs =
( 4.0 furlongs ) ( 201.168 m furlong )
5.0292 m rod
= 160 rods,
(b) and that distance in chains to be
d=
( 4.0 furlongs ) ( 201.168 m furlong )
20.117 m cha
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