HW 8 Solutions
Chapter 15 Problem 7
In exercising, a weight lifter loses 0.150 kg of water through evaporation, the heat required to
evaporate the water coming from the weight lier's body. The work done in lifting weights is
1.40 x 105 J. (a) Assuming tha
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
2. A nano
1. Each atom has a mass of m = M/NA, where M is the molar mass and NA is the
Avogadro constant. The molar mass of arsenic is 74.9 g/mol or 74.9103 kg/mol.
Therefore, 7.50 1024 arsenic atoms have a total mass of
(7.50 1024) (74.9 103 kg/mol)/(6.
1. (a) All the energy in the circuit resides in the capacitor when it has its maximum
charge. The current is then zero. If Q is the maximum charge on the capacitor, then the
total energy is
2.90 106 C
117 106 J.
2C 2 3.60 106 F
1. THINK The magnetic force on a charged parti014cle is given by FB qv B, where
v is the velocity of the charged particle and B is the magnetic field.
EXPRESS The magnitude of the magnetic force on the proton (of charge +e) is
FB evB sin , wher
1. We use
Bn 0 to obtain
B 6 Bn 1Wb 2 Wb 3Wb 4 Wb 5 Wb 3Wb .
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. Since the net flux must be ze
1. The fact that wave W2 reflects two additional times has no substantive effect on the
calculations, since two reflections amount to a 2(/2) = phase difference, which is
effectively not a phase difference at all. The substantive difference bet
1. Since , we find f is equal to
c c (3.0 10 m/s)(0.0100 10 m)
7.49 109 Hz.
(632.8 10 m)
2. (a) The frequency of the radiation is
3.0 108 m / s
4.7 103 Hz.
(10 10 )(6.4 10 m)
(b) The period of the radiation is
1. The bird is a distance d2 in front of the mirror; the plane of its image is that same
distance d2 behind the mirror. The lateral distance between you and the bird is d3 = 5.00
m. We denote the distance from the camera to the mirror as d1, an
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 0 .
With r = 20 ft = 6.10 m, we have
2 b mg
T m A 100 A
g 3.3 10
T 3.3 T.
(b) This is about one-sixt
1. The flux B BA cos does not change as the loop is rotated. Faradays law only
leads to a nonzero induced emf when the flux is changing, so the result in this instance is
2. Using Faradays law, the induced emf is
d r 2
1. THINK If the expansion of the gas is reversible and isothermal, then theres no change
in internal energy. However, if the process is reversible and adiabatic, then there would
be no change in entropy.
EXPRESS Since the gas is ideal, its pres
1. THINK After the transfer, the charges on the two spheres are Q q and q.
EXPRESS The magnitude of the electrostatic force between two charges q1 and q2
separated by a distance r is given by the Coulombs law (see Eq. 21-1):
where k 1
1. (a) The capacitance of the system is
(b) The capacitance is independent of q; it is still 3.5 pF.
(c) The potential difference becomes
q 200 pC
C 35 pF
2. Charge flows until the potential difference acr
1. We note that the symbol q2 is used in the problem statement to mean the absolute value
of the negative charge that resides on the larger shell. The following sketch is for q1 q2 .
The following two sketches are for the cases q1 > q2 (left fi
1. THINK This exercise deals with electric flux through a square surface.
EXPRESS The vector area A and the electric field E are shown on the diagram below.
The electric flux through the surface is given by E A EA cos .
EXPRESS The angle betwee
1. THINK Ampere is the SI unit for current. An ampere is one coulomb per second.
EXPRESS To calculate the total charge through the circuit, we note that 1 A 1 C/s and
1 h 3600 s.
ANALYZE (a) Thus,
84 A h 84
IJ FG 3600 s IJ 3.0 10
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
1. THINK The circuit consists of two batteries and two resistors. We apply Kirchhoffs
loop rule to solve for the current.
EXPRESS Let i be the current in the circuit and take it to be positive if it is to the left in
R1. Kirchhoffs loop rule gi
1. (a) With E = hc/min = 1240 eVnm/min = 0.6 eV, we obtain = 2.1 103 nm = 2.1
(b) It is in the infrared region.
me v 2 E photon
and solve for v:
me c 2
2.998 108 m/s
2 1240 eV nm
590 nm 51
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
The proper time interval is measured by a clock at rest relative to the muon. Specifically,
t0 = 2.2000 s.
1. (a) We use Eq. 36-3 to calculate the separation between the first (m1 = 1) and fifth
(m2 5) minima:
y D sin D
m2 m1 .
Solving for the slit width, we obtain
400 mm 550 106 mm 5 1
D m2 m1
2.5 mm .
Midterm Exam II set A (29:081)
Mar 07, 2012
Closed book, all questions carry equal credit
No calculators or formula sheet allowed
1. If a certain car, going with speed v1, rounds a level curve with a radius R1, it is just on the
verge of skidding. If its
MATH:1850 Calculus I Final Exam Study Guide
Midterm 1 exam will be held on Friday, February 10, 2017 in class. The test will cover
Chapter 1 (1.1, 1.2, 1.3, 1.5) and Chapter 2 (2.2, 2.3, 2.4, 2.5, 2.6).
Formal definition of a limit: The limit of function
Homework Z cfw_a \qul Ms
Chapter 03, Problem 038
For the following three vectors, what is 2 C -(2 A x B )?
A A A
A = 3.003' + 3.00 - 4.001:
A A A
B = 3.001' + 4.00 + 4.00%:
:2 = 8.001' - 3.00;
Number .395 Units lThis answer has no units C
Hewework S Sold-1c k3
You drop a 2.00 kg book to a friend who stands on the ground at distance D = 14.0 m below. If your friend's
outstretched hands are at distance d = 1.70 m above the ground (see the figure), (a) how much work Wg does
Chapter 01, Problem 008
Harvard Bridge, which connects MIT with its fraternities across the Charles River, has a length of 364.4 Smoots
plus one ear. The units of one Smoot is based on the length of Oliver Reed Smoot, Jr., class of 1962, who was
HomionLC E $50410 as
Chapter 09, Problem 028
Flying Circus of Physics
In tae-kwondo, a hand is slammed down onto a target at a speed of 12 m/s and comes to a stop during the
2.0 ms collision. Assume that during the impact the hand is independent of the
A 3.6 kg block is pushed along a horizontal floor by a force F of magnitude 21 N at a downward angle
Chapter 06, Problem 009
9 = 40. The coefcient of kinetic friction between the block and the oor is 0.25. Calculate the magnitudes of