ELECTRIC CHARGES AND FORCES
25.1. Model: Use the charge model. Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred to the other because they are relatively free to move. Protons, on the other han

THEELECTRIC POTENTIAL
29.1. Model: The mechanical energy of the proton is conserved. A parallel plate capacitor has a uniform
electric field. Visualize:
After Before
*
. .
,
v
=o
E`
*
I
I
0
1.o
x
2.0
The figure shows the before-and-afte

THEELECTRIC FIELD
26.1. Model: The electric field is that of the two charges placed on the y-axis.
Visualize: Please refer to Figure Ex26.1. We denote the upper charge by q1 and the lower charge by q2. Because
both the charges are positive, their el

17.1. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, Eth =
Kmicro. Solve: The number of atoms is
N=
M 0.0020 kg = = 3.01 10 23 m 6.64 10 -27 kg
Because helium atoms have an atomic mass number A

16.1. Solve: The mass of lead mPb = Pb VPb = (11,300 kg m 3 )(2.0 m 3 ) = 22,600 kg . For water to have the
same mass its volume must be
Vwater =
mwater 22,600 kg = = 22.6 m 3 water 1000 kg m 3
16.2. Solve: The volume of the uranium nucleus is
V

14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, hence
T= 1 1 = = 2.27 10 -3 s = 2.27 ms f 440 Hz
14.2. Solve: Your pulse or heart beat is 75 beats per minute. The frequency of your hear

13.1. Model: The crankshaft is a rotating rigid body.
Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s, maintains a constant angular velocity for 2 s until t = 4 s, and then speeds up

12.1.
Solve: (b)
Model: Model the sun (s), the earth (e), and the moon (m) as spherical. (a)
Fs on e =
Gms me (6.67 10 -11 N m 2 / kg 2 )(1.99 10 30 kg)(5.98 10 24 kg) = 3.53 10 22 N = (1.50 1011 m ) 2 rs2 e -
Fm on e =
GMm Me (6.67 10 -1

10.1. Model: We will use the particle model for the bullet (B) and the bowling ball (BB).
Visualize:
Solve:
For the bullet,
KB =
For the bowling ball,
1 1 2 mB vB = (0.01 kg)(500 m /s) 2 = 1250 J 2 2 1 1 2 mBB vBB = (10 kg)(10 m / s) 2 = 500 J 2

7.1. Solve: (a) From t = 0 s to t = 1 s the particle rotates clockwise from the angular position +4 rad to -2 rad. Therefore, = -2 - ( +4 ) = -6 rad in one sec, or = -6 rad s . From t = 1 s to t = 2 s, = 0 rad/s. From t = 2 s to t = 4 s the partic

22.1. Visualize: Please refer to Figure Ex22.1.
Solve: (a)
(b) The initial light pattern is a double-slit interference pattern. It is centered behind the midpoint of the slits. The slight decrease in intensity going outward from the middle indicates

24.1. Model: Balmer's formula predicts a series of spectral lines in the hydrogen spectrum.
Solve: Substituting into the formula for the Balmer series,
=
91.18 nm 91.18 nm = = 410.3 nm 1 1 1 1 - 2 - 2 2 22 n 2 6
where n = 3, 4, 5, 6, . and wher

1.1.
Solve:
1.2.
Solve:
Solve: (a) The basic idea of the particle model is that we will treat an object as if all its mass is concentrated into a single point. The size and shape of the object will not be considered. This is a reasonable approxim

2.1.
Solve:
Model: The car is represented by the particle model as a dot. (a) Time t (s) Position x (m) 0 1200 1 975 2 825 3 750 4 700 5 650 6 600 7 500 8 300 9 0
(b)
2.2. Solve:
Diagram (a) (b) (c)
Position Negative Negative Positive
Velocity

3.1. Solve: (a) If one component of the vector is zero, then the other component must not be zero (unless the whole vector is zero). Thus the magnitude of the vector will be the value of the other component. For example, if Ax = 0 m and Ay = 5 m, the

4.1. Solve: A force is basically a push or a pull on an object. There are five basic characteristics of forces. (i) A force has an agent that is the direct and immediate source of the push or pull. (ii) Most forces are contact forces that occur at a

5.1.
Model: We can assume that the ring is a single massless particle in static equilibrium. Visualize:
Solve:
Written in component form, Newton's first law is
( Fnet ) x = Fx = T1x + T2 x + T3 x = 0 N
T1 x = - T1
T1y = 0 N Using Newton's first l

Additional Examples #33
Whereas lenses form images through refraction, mirrors form images through reflection. Incident
electromagnetic waves parallel to the optical axis of a concave spherical mirror are reflected
toward a point. Incident electromagnetic

Additional Examples #32
A lens is an optical device that refracts an electromagnetic wave; most lenses rely upon curved
surfaces to impart further refraction of an incident electromagnetic wave. The optical axis of a
lens is the axis passing through the c

6.1. Model: We will assume motion under constant-acceleration kinematics in a plane.
Visualize:
Instead of working with the components of position, velocity, and acceleration in the x and y directions, we will use the kinematic equations in vector f

3.1. Solve: (a) If one component of the vector is zero, then the other component must not be zero (unless the whole vector is zero). Thus the magnitude of the vector will be the value of the other component. For example, if Ax = 0 m and Ay = 5 m, the

FUNDAMENTALS OF CIRCUITS
31.1. Solve: From Table 30.1, the resistivity of carbon is p = 3.5 x
of lead from a mechanical pencil is
R m. From Equation 31.3, the resistance
= 5.5 R
p~ p~ R=-=-= A m '
(3.5 x lo-'
R m)(0.06 m)
n(0.35~10-'m)'
31.2.

Solve: (a) The momentum p = mv = (1500 kg)(10 m /s) = 1.5 10 4 kg m /s . (b) The momentum p = mv = (0.2 kg)( 40 m /s) = 8.0 kg m /s .
9.1. Model: Model the car and the baseball as particles.
9.2. Model: Model the bicycle and its rider as a particl

POTENTIAL AND FIELD
30.1. Solve: The potential difference AV between two points in space is
9
AV = V(xf) - V(x,) = - I E , dx
x,
where x is the position along a line from point i to point f. When the electric field is uniform,
xr
AV = - E x j d .

8.1. Visualize:
Solve: Figure (i) shows a weightlifter (WL) holding a heavy barbell (BB) across his shoulders. He is standing on a rough surface (S) that is a part of the earth (E). We distinguish between the surface (S), which exerts a contact forc

CURRENT AND CONDUCTIVITY
0 28.1. Solve: The wire's cross-sectional area is A = mz = ~ ( 1 . x 10" m)' = 3.1415 x lo4 m2, and the electron current through this wire is 2.0 X loi9 s-l . Using Table 28.1 for the electron density of iron and Equation 28

5.1.
Model: We can assume that the ring is a single massless particle in static equilibrium. Visualize:
Solve:
Written in component form, Newton's first law is
( Fnet ) x = Fx = T1x + T2 x + T3 x = 0 N
T1 x = - T1
T1y = 0 N Using Newton's first l

GAUSS'S LAW
27.1. V i a l i e :
As discussed in Section 27.1, the symmetry of the electric field must match the symmetry of the charge distribution. In particular, the electric field of a cylindrically symmetric charge distribution cannot have a co

4.1. Solve: A force is basically a push or a pull on an object. There are five basic characteristics of forces. (i) A force has an agent that is the direct and immediate source of the push or pull. (ii) Most forces are contact forces that occur at a

11.1. Visualize:r Please refer to Figure Ex11.1. r
Solve: (b) (c)
(a) A B = AB cos = ( 4)(5)cos 40 = 15.3. r r C D = CD cos = (2)( 4)cos120 = -4.0. r r E F = EF cos = (3)( 4)cos 90 = 0.
11.2. Visualize:r Please refer to Figure Ex11.2. r
Solve