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 pre
Your Name_Section_
HOMEWORK # 1 - 8.01 MIT - Prof. Kowalski
Due 4:00PM Thursday, Sept. 11, 2003
Topics: Dimensions, Units, and One dimensional motion
1. Sunset at the Equator
Sunset is defined as the
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 relat
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
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. 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 i
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 mag
1. According to Eq. 39-4 En L 2. As a consequence, the new energy level E'n satisfies
En L = En L
FG IJ = FG L IJ H K H L K
-2
2
=
1 , 2
which gives L = 2 L. Thus, the ratio is L / L = 2 = 1.41.
2. (a
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
1. From the time dilation equation t = t0 (where t0 is the proper time interval,
= 1 / 1 - 2 , and = v/c), we obtain
= 1-
FG t IJ . H t K
2 0
The proper time interval is measured by a clock at rest
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
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
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 = 13
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 elec
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
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
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
1. (a) With a understood to mean the magnitude of acceleration, Newton's second and third laws lead to m2 a2 = m1a1
c6.3 10 kghc7.0 m s h = 4.9 10 m =
-7 2 2
-7
9.0 m s
2
kg.
(b) The magnitude of the
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
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 pe
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
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.
1. With speed v = 11200 m/s, we find
K= 1 2 1 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
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 c
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
HW Solutions # 14 - 8.01 MIT - Prof.
Kowalski
Relativity. 1) 37.20
Relativistic Relative Velocity
Write dierential form of Lorentz transformation: x = (x ut) t = (t ux/c2 ) 1 = 1 u2 /c2 which is: dx =
HW Solutions # 13 - 8.01 MIT - Prof.
Kowalski
Harmonic Oscillators and Relative Motion.
1)
Vibration Isolation System
a) Lets denote the table displacement from equilibrium with Y . The Newtons second
HW Solutions # 12 - 8.01 MIT - Prof.
Kowalski
Universal Gravity and Harmonic Oscillators.
1)
Whither the Moon?
a) TIDES: The water (and to some extent the liquid core) of the earth bulge out toward an
HW Solutions # 11 - 8.01 MIT - Prof.
Kowalski
Universal Gravity. 1) 12.23
Escaping From Asteroid
Please study example 12.5 "from the earth to the moon".
a) The escape velocity derived in the example (
HW Solutions # 10 - 8.01 MIT - Prof.
Kowalski
Angular Energy and Angular Momentum.
1) 10.70
Please refer to figure 10.54 p.399. The general strategy to solve this problem is to gure out the torque aro