Chapter 14
1. The pressure increase is the applied force divided by the area: p = F/A = F/r2, where r is the radius of the piston. Thus p = (42 N)/(0.011 m)2 = 1.1 105 Pa. This is equivalent to 1.1 atm. 2. We note that the container is cylindrical, the im
Chapter 33
1. (a) From Fig. 33-2 we find the smaller wavelength in question to be about 515 nm. (b) Similarly, the larger wavelength is approximately 610 nm. (c) From Fig. 33-2 the wavelength at which the eye is most sensitive is about 555 nm. (d) Using t
Chapter 32
1. We use
6 n=1
Bn = 0 to obtain
5 n=1
B6 = -
Bn = - ( -1Wb + 2 Wb - 3 Wb + 4 Wb - 5 Wb ) = +3 Wb .
2. (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.
Chapter 31
1. (a) The period is T = 4(1.50 s) = 6.00 s. (b) The frequency is the reciprocal of the period: f = 1 1 = = 167 105 Hz. . T 6.00s
(c) The magnetic energy does not depend on the direction of the current (since UB i2), so this will occur after on
Chapter 30
1. (a) The magnitude of the emf is
=
d B d = 6.0t 2 + 7.0t = 12t + 7.0 = 12 2.0 + 7.0 = 31 mV. dt dt
c
h
bg
(b) Appealing to Lenz's law (especially Fig. 30-5(a) we see that the current flow in the loop is clockwise. Thus, the current is to lef
Chapter 29
1. (a) The field due to the wire, at a point 8.0 cm from the wire, must be 39 T and must be directed due south. Since B = 0i 2 r ,
-6 2 rB 2 0.080 m 39 10 T i= = = 16 A. 0 4 10 -7 T m A
b
g c
h
(b) The current must be from west to east to produ
Chapter 28
1. (a) The force on the electron is r r r FB = qv B = q vx ^ + v y ^ i j
= -1.6 10-19 C 2.0 106 m s ( -0.15 T ) - 3.0 106 m s ( 0.030 T ) -14 ^ = 6.2 10 N k.
( (
(
)
) (
) ( B ^i + B j ) = q ( v B
r
x y x
y
^ - v y Bx ) k
)
(
)
r r Thus, the ma
Chapter 27
1. (a) The energy transferred is
2t (2.0 V) 2 (2.0 min) (60 s / min) U = Pt = = = 80 J . r+R 1.0 + 5.0
(b) The amount of thermal energy generated is
F I F 2.0 V I (5.0 ) (2.0 min) (60 s / min) = 67 J. U = i Rt = G J Rt = G H+ R K H + 5.0 J r
Chapter 26
1. (a) The charge that passes through any cross section is the product of the current and time. Since t = 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 t
Chapter 25
1. (a) The capacitance of the system is C= q 70 pC = = 35 pF. . V 20 V
(b) The capacitance is independent of q; it is still 3.5 pF. (c) The potential difference becomes V = q 200 pC = = 57 V. C 35 pF .
2. Charge flows until the potential differ
Chapter 24
1. If the electric potential is zero at infinity then at the surface of a uniformly charged sphere it is V = q/40R, where q is the charge on the sphere and R is the sphere radius. Thus q = 40RV and the number of electrons is q 4 0 R V ( 1.0 10-
Chapter 23
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 2 = E A = EA cos = ( 1800 N C ) ( 3.2 10 -3 m ) cos145 = -1.5 10 -2 N m 2 C.
2. We
Chapter 17
1. (a) When the speed is constant, we have v = d/t where v = 343 m/s is assumed. Therefore, with t = 15/2 s being the time for sound to travel to the far wall we obtain d = (343 m/s) (15/2 s) which yields a distance of 2.6 km. (b) Just as the
1
Chapter 16
1. (a) The angular wave number is k = 2 2 = = 3.49 m -1. 1.80 m
(b) The speed of the wave is v = f =
( 1.80 m ) ( 110 rad s ) = = 31.5 m s. 2 2
2. The distance d between the beetle and the scorpion is related to the transverse speed vt and lon
Chapter 15
1. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is am = 2xm, where is the angular frequency ( = 2f since there are 2 radians in one cycle). Therefore, in this circumstance, we obtain am = 2 x
Chapter 13
1. The magnitude of the force of one particle on the other is given by F = Gm1m2/r2, where m1 and m2 are the masses, r is their separation, and G is the universal gravitational constant. We solve for r: Gm1m2 r= = F
( 6.67
10-11 N m 2 / kg 2 )
Chapter 12
1. (a) The center of mass is given by xcom = 0 + 0 + 0 + (m)(2.00 m) + ( m)(2.00 m) + ( m)(2.00 m) = 1.00 m. 6m
(b) Similarly, we have ycom = 0 + (m)(2.00 m) + (m)(4.00 m) + ( m)(4.00 m) + ( m)(2.00 m) + 0 = 2.00 m. 6m
(c) Using Eq. 12-14 and n
Chapter 11
1. The velocity of the car is a constant r ^ v = + ( 80 km/h ) (1000 m/km)(1 h/3600 s) ^ = (+22 m s)i, i and the radius of the wheel is r = 0.66/2 = 0.33 m. (a) In the car's reference frame (where the lady perceives herself to be at rest) the r
Chapter 10
1. The problem asks us to assume vcom and are constant. For consistency of units, we write 5280 ft mi vcom = 85 mi h = 7480 ft min . 60 min h
b
F g G H
I J K
Thus, with x = 60 ft , the time of flight is
t = x vcom = (60 ft) /(7480 ft/min) = 0.0
Chapter 9
1. We use Eq. 9-5 to solve for ( x3 , y3 ). (a) The x coordinates of the system's center of mass is: xcom = m1 x1 + m2 x2 + m3 x3 (2.00 kg)(-1.20 m) + ( 4.00 kg ) ( 0.600 m ) + ( 3.00 kg ) x3 = m1 + m2 + m3 2.00 kg + 4.00 kg + 3.00 kg
= -0.500 m
Chapter 8
1. (a) Noting the vertical displacement is 10.0 m 1.50 m = 8.50 m downward (same that direction as Fg ), Eq. 7-12 yields Wg = mgd cos = (2.00 kg)(9.80 m/s 2 )(8.50 m) cos 0 = 167 J. (b) One approach (which is fairly trivial) is to use Eq. 8-1, b
Chapter 7
1. (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 negative sign indicates that kinetic energy is lost. (b) The energy l
Chapter 6
1. We do not consider the possibility that the bureau might tip, and treat this as a purely horizontal motion problem (with the person's push F in the +x direction). Applying Newton's second law to the x and y axes, we obtain F - f s , max = ma
Chapter 5
1. We apply Newton's second law (specifically, Eq. 5-2). (a) We find the x component of the force is Fx = max = ma cos 20.0 = ( 1.00kg ) ( 2.00m/s 2 ) cos 20.0 = 1.88N. (b) The y component of the force is Fy = ma y = ma sin 20.0 = ( 1.0kg ) ( 2.
Chapter 4
1. The initial position vector ro satisfies r - ro = r , which results in
^ ^ ^ ^ ^ ro = r - r = (3.0^ - 4.0k)m - (2.0i - 3.0^ + 6.0 k)m = (-2.0 m) i + (6.0 m) ^ + (- 10 m) k . j j j
^ 2. (a) The position vector, according to Eq. 4-1, is r =
Chapter 3
r 1. A vector a can be represented in the magnitude-angle notation (a, ), where
2 a = ax + a 2 y
is the magnitude and a = tan - 1 y a x r is the angle a makes with the positive x axis. (a) Given Ax = -25.0 m and Ay = 40.0 m, A = (- 25.0 m) 2 + (
Chapter 1
1. The metric prefixes (micro, pico, nano, .) are given for ready reference on the inside front cover of the textbook (see also Table 12). (a) Since 1 km = 1 103 m and 1 m = 1 106 m, 1km = 103 m = 103 m 106 m m = 109 m. The given measurement is
Chapter 44
1. The total rest energy of the electron-positron pair is E = me c 2 + me c 2 = 2me c 2 = 2(0.511 MeV) = 1.022 MeV . With two gamma-ray photons produced in the annihilation process, the wavelength of each photon is (using hc = 1240 eV nm )
=
hc
Chapter 43
1. If MCr is the mass of a 52Cr nucleus and MMg is the mass of a 26Mg nucleus, then the disintegration energy is Q = (MCr 2MMg)c2 = [51.94051 u 2(25.98259 u)](931.5 MeV/u) = 23.0 MeV. 2. Adapting Eq. 42-21, there are N Pu = M sam 1000 g NA = (
Chapter 42
1. Kinetic energy (we use the classical formula since v is much less than c) is converted into potential energy (see Eq. 24-43). From Appendix F or G, we find Z = 3 for Lithium and Z = 90 for Thorium; the charges on those nuclei are therefore 3