Chapter 20 Student Solutions Manual 5. (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 = p dV = n
Chapter 8 Student Solutions Manual
1 5. The potential energy stored by the spring is given by U = 2 kx 2 , where k is the spring constant and x is the displacement of the end of the spring from its position when the spring is in equilibrium. Thus
k
Chapter 6 Student Solutions Manual 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
Chapter 5 Student Solutions Manual 5. We denote the two forces F1 and F2 . According to Newton's second law, F1 + F2 = ma , so F2 = ma - F1 . (a) In unit vector notation F1 = 20.0 N i and
a = - (12.0 sin 30.0 m/s 2 ) ^ - (12.0 cos 30.0 m/s 2 ) ^ = -
Chapter 4 Student Solutions Manual 11. We apply Eq. 4-10 and Eq. 4-16. (a) Taking the derivative of the position vector with respect to time, we have, in SI units (m/s),
v= d ^ ^ (i + 4t 2 ^ + t k) = 8t ^ + k . j j ^ dt
(b) Taking another derivativ
Chapter 3 Student Solutions Manual 1. A vector a can be represented in the magnitude-angle notation (a, ), where
2 2 a = ax + a y
is the magnitude and
= tan - 1
is the angle a makes with the positive x axis.
ay ax
(a) Given Ax = -25.0 m a
Chapter 2 Student Solutions Manual 1. We use Eq. 2-2 and Eq. 2-3. During a time tc when the velocity remains a positive constant, speed is equivalent to velocity, and distance is equivalent to displacement, with x = v tc. (a) During the first part o
Chapter 1 Student Solutions Manual 3. 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
Chapter 1:
MEASUREMENT
1. The SI standard of time is based on: A. the daily rotation of the earth B. the frequency of light emitted by Kr86 C. the yearly revolution of the earth about the sun D. a precision pendulum clock E. none of these Ans: E 2.
Chapter 9 Student Solutions Manual 15. We need to find the coordinates of the point where the shell explodes and the velocity of the fragment that does not fall straight down. The coordinate origin is at the firing point, the +x axis is rightward, a
Chapter 10 Student Solutions Manual 13. We take t = 0 at the start of the interval and take the sense of rotation as positive. 1 Then at the end of the t = 4.0 s interval, the angular displacement is = 0t + 2 t 2 . We solve for the angular velocit
Chapter 19 Student Solutions Manual 7. (a) In solving pV = nRT for n, we first convert the temperature to the Kelvin scale: T = (40.0 + 273.15) K = 313.15 K . And we convert the volume to SI units: 1000 cm3 = 1000 106 m3. Now, according to the idea
Chapter 18 Student Solutions Manual 8. The change in length for the aluminum pole is
=
0
A1T = (33m)(23 10 -6 / C)(15 C) = 0.011m.
15. If Vc is the original volume of the cup, a is the coefficient of linear expansion of aluminum, and T is the t
Chapter 17 Student Solutions Manual 5. Let tf be the time for the stone to fall to the water and ts be the time for the sound of the splash to travel from the water to the top of the well. Then, the total time elapsed from dropping the stone to hear
Chapter 16 Student Solutions Manual 15. The wave speed v is given by v = , where is the tension in the rope and is the linear mass density of the rope. The linear mass density is the mass per unit length of rope:
= m/L = (0.0600 kg)/(2.00 m) =
Chapter 15 Student Solutions Manual 3. (a) The amplitude is half the range of the displacement, or xm = 1.0 mm. (b) The maximum speed vm is related to the amplitude xm by vm = xm, where is the angular frequency. Since = 2f, where f is the frequenc
Chapter 14 Student Solutions Manual 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. 3. The air inside pu
Chapter 13 Student Solutions Manual 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= =
Chapter 12 Student Solutions Manual 5. Three forces act on the sphere: the tension force T of the rope (acting along the rope), the force of the wall FN (acting horizontally away from the wall), and the force of gravity mg (acting downward). Since t
Chapter 11 Student Solutions Manual 5. By Eq. 10-52, the work required to stop the hoop is the negative of the initial kinetic 1 1 energy of the hoop. The initial kinetic energy is K = 2 I 2 + 2 mv 2 (Eq. 11-5), where I = mR2 is its rotational inert
Physics 171.101
Second Exam Nov 8, 2005 As a reminder that we are striving to establish a culture of integrity at JHU, please sign the following statement and return this with your blue book. The exam will be posted on our class web site after all ex
Physics 171.101
First Exam Oct. 11, 2005 As a reminder that we are striving to establish a culture of integrity at JHU, please sign the following statement and return this with your blue book. The exam will be posted on our class web site after all e
1. The x and the y components of a vector a lying on the xy plane are given by
ax = a cos , 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
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
2r
.
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
1. (a) Eq. 28-3 leads to 6.50 10-17 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 1 2 1 mv = 167 10-27 kg 4.00 105 m s = 134 10-16 J. . . 2 2
c
hc
h
This is
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 magnitu
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.
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 10-3 m cos145 = -1.5 10-2 N m 2 C.
2
(
)
2. We u
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 figu