. motion of medium (SHM): Wave Equation (1D):
S u m m a W S h eet Waves nd a deswed part Of wave (wave does not tranSPOrt matter) , holds for all travelling waves & superpositions gall/3:211:33
(or comparing two waves):
. . _. _ Ad) Wm, 2 2 fsta 5 same
p
Physics 270 - HW#1 Solutions
to balance off net inward
/, = 1 / find what Fpu |has to be
7 _ 7
;
force (technically F
5: - IL I: : 0 must be ever-so- slightly
- larger than that, to lift
._ - l' ff
ll _ f;- F; Id 0 )
Box is rst sealed while on mountain
c = specific heat cap: J/(kg K)
Thermo SU mmary sheet Heat Capacity: Q=chT or nCAT Simonarhemm J/(moleK) Misc Q exchange:
(M=molar mass)
depends on: material phase process temperature
particularly c gases only how many f
(depends on M) (CP> CV)
14.) In a SHM system, the object is 85% of its way to maximum distance (or angle) from equilibrium at t=0. Please
find the smallest positive (and negative) phase angles for our standard SHM equation, if the position at this time is
_ while the direction o
Simple* Harmonic Motion
*T dependent on inherent system properties,
not initial conditions _
mass on Sprlng:
Fig.4.
from spring
restoring agent.
must grow linearly w/
displace from equil
if to cause HM
.causes motion. F
Newtons 2nd Law
combine _ 2
two
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 note that the sum of superscripts (mass numbers A) must
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 ) (1.60 10 C )( 8.99 10 V m/C ) = = = 4 0 K 4 0 K 5.30 106
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 is also the number of conduction electrons per unit volu
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 magnetic quantum number m ranges from - to + . For = 1 , t
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) The ground-state energy is
( 6.63 10 J s ) h2 E1 = n2
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. The wavelength is related to the frequency by f = c,
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 relative to the muon. Specifically, t0 = 2.2000 s. We a
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 sides must be negative and exactly cancel the total of
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) This is about one
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 (1.34 10 16 J) / (1.6
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 magnitude of
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. Thus,
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 use = E A , whe
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 figure) an
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 (only) force on particle 1 is
q q q F = m1a1 = k 1 2 2
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
W=
V2 V1
p dV = n RT
V2 V1
dV V = n RT ln 2 . V V1
We s
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 period:
f = 1 1 = = 1.47 Hz. T 0.680 s
(c) A sinusoidal w
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 by poA, where po is the pressure outside. The magnitud
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 related to its energy by p = E/c, the particles have the sam
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 kg 15 103 m/s 2 2
(
)(
)
2
= -5 1014 J ,
or | K |= 5 1
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 = 2.5 m. (b)
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 d =
( 4.0 furlongs )( 201.168 m furlong )
20.117 m chai
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 furlongs