Ph1c Practical Quiz Solutions 2
Spring 2014
2.1
2.1.a
(5 points)
(1 point)
Using the Faradays Law:
L = a2 B0 cos t;
R = 0
2
IL =
2.1.b
VL
1 dL
a wB0
=
=
sin t
R
R dt
R
VR
1 dR
IR =
=
=0
R
R dt
(1 point)
Since BR = 0 and IR = 0, The net force on the left a
June 6, 2014
PHYSICS 1c Prac FINAL EXAM
This is an OPEN BOOK quiz with the following limitations: You may consult only
your textbook (Serway & Jewett), notes that you have taken in recitation or lecture,
course handouts, your own homework, and the posted
Physics 1c Practical, Spring 2015
Homework 8 Solutions
Serway 39.32
ux = velocity of other jet in frame of jet
ux = velocity of other jet in frame of galaxy center
v = speed of galaxy center in frame of jet
From equation 39.16
ux =
ux v
0.75c 0.75c
=
= 0.
ACM 95/100a
Week 3 lecture notes
Dan Meiron
Caltech
January 18, 2017
D. Meiron (Caltech)
ACM 95/100a - Methods of Applied Mathematics for the Physical
January
Sciences
18, 2017
1 / 49
Bounding complex integrals
We often have to produce bounding values for
ACM 95/100a
Week 1 lecture notes
Dan Meiron
Caltech
January 4, 2017
D. Meiron (Caltech)
ACM 95/100a - Methods of Applied Mathematics for the Physical
January
Sciences
4, 2017
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Introduction
Welcome to ACM 95/100a
Instructor: Dan Meiron
Office: 343 Gu
ACM 95/100a
Week 2 lecture notes
Dan Meiron
Caltech
January 15, 2017
D. Meiron (Caltech)
ACM 95/100a - Methods of Applied Mathematics for the Physical
January
Sciences
15, 2017
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Branch points
In defining the Riemann surface of a complex function the
ACM 95/100a
Week 4 lecture notes
Dan Meiron
Caltech
January 26, 2017
D. Meiron (Caltech)
ACM 95/100a - Methods of Applied Mathematics for the Physical
January
Sciences
26, 2017
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Using the generalized Cauchy formula to evaluate
contour integrals
As a
Ph1c Practical Quiz 1 Solution
April 26, 2014
Problem 1 (6 points)
L
2v
L
2v
Fext x
(b)
t
F
ext
N S
t
F
ext
N
P = I 2R
(c)
1.1
S
L
2v
L
2v
N S
Iring
(a)
L
2v
L
2v
t
(a) (2 points)
When the ring is far from the solenoid, the ux is almost zero. It is a fair
and let the device of interest be modeled by a square loop of wire of area A located a distance R
from the power line. R A1/2 so that the magnetic eld due do I(t) can be approximated as a
constant at any given time over the whole area A. Assume that the p
Ph1c Practical Quiz 3 Solutions
Spring 2014
3.1
3.1.a
(10 points)
(2 points)
IL (t) =
1
L
V (t)dt =
IC (t) = C
3.1.b
d
V (t) = V0 C cos t
dt
(2 point)
Itot = IL + IC = V0 C
1
cos t = 0
L
1
1
=
0
L
LC
C =
3.1.c
V0
cos t
L
(2 points)
1 2
1 V02
LIL (t)
Ph1c Practical Quiz Solution 4
Spring 2014
4.1
4.1.a
(8 points)
(1 point)
The capacitance of a parallel plate capacitor is
0 A
D
C=
For circular plate capacitor, we simply subsitute
A = R2
Obtaining
C=
4.1.b
0 R2
D
(2 points)
The electric eld energy densi
Physics 1c Practical, Spring 2015
Quiz 2 Solutions
May, 2015
Problem 1 (6 points)
(a) (2 points)
According to Faradays law, the E.M.F due to the change of magnetic field is
E=
d
;
dt
)E = a2 cos
~ S
~ = a2 B cos
=B
(1)
dB
dt
(2)
The self-inductance of t
Physics 1c Practical, Spring 2015
Quiz 3 Solutions
May 19, 2015
1
(a)
(8 points)
(3 points)
We know that ZC =
1
i!C
and ZL = i!L and ZR = R. So possible black box impedances are:
i
!C
= i!L
Zbb, C =
Zbb, L
i
Zbb, series = i!L
!C
1
Zbb, parallel =
+ i!C
i!
Physics 1c practical, 2015
Serway 34.22
a
P
energy
600 103 Wh
=
=
area
t area
(30 d)(13.0 m)(9.50 m)
1d
24 h
= 6.75 W/m2
(1)
b
The car uses gasoline at the rate of(55mi/h) 25galmi . Its rate of energy conversion is P = 44.0
gal
1h
4
106 J/kg 2.54kg
(55mi
Physics 1c Practical, Spring 2015
Homework 7 Solutions
Serway 37.4
ybright =
L
d m.
For m = 1, =
ybright .d
L
=
(3.4103 ).(0.5103 )
3.3
= 515 nm
Serway 37.13
Note, with the conditions given, the small-angle approximation does not work well. That is,
sin ,
Ph1c Practical Homework Solutions 3
Spring 2014
3.1
3.1.a
Serway 31.80 (5 points)
(3.5 points)
We divide up the disk into many concentric rings. Consider the power dissipated in dierent rings and then
sum them up. By Faradays law of induction, we work out
Ph1c Practical Homework Solutions 8
Spring 2014
8.1
Serway 39.7 (5 points)
Time is dilated by a factor 2, which means = 2.
= 1/ 1 v 2 /c2
v=c
8.2
8.2.a
v = c 1 1/ 2
1 1/22 = c 3/2 = 0.866c = 2.60 108 m/s
Serway 39.8 (5 points)
(3 points)
The lifetime of
Ph1c Practical Homework Solutions 10
Spring 2014
10.1
Serway 39.56 (5 points)
By conservation of energy, the photon energy must be at least the sum of the rest energies of the created
particles:
E me+ c2 + me c2 = 2me c2 = 2(0.511 MeV/c2 ) c2 ,
E 1.02 MeV
Ph1c Practical Homework Solutions 9
Spring 2014
9.1
Serway 39.33 (5 points)
This is simply an application of the relativistic velocity addition formula:
u0 =
K
1
uK v
,
uK v/c2
where uK is the speed of the (enemy) Klingon ship in the Earths frame (0.800c)
Ph1c Practical Homework Solutions 5
Spring 2014
5.1
5.1.a
Serway 33.71 (5 points)
(2.5 points)
For high frequencies, the inductor stops the signal and the capacitor lets the signal pass unimpeded. The
only relevant circuit element is the top resistor:
I =
Ph1c Practical Homework Solutions 7
Spring 2014
7.1
Serway 34.17 (5 points)
The distance between two consecutive antinodes is /2.
= 12cm 5%
Therefore, the speed of the microwave c = f is,
c = f = 2.94 108 m/s 5%.
7.2
Serway 34.79 (5 points)
In the deriva
Ph1c Practical Homework Solutions 4
Spring 2014
4.1
4.1.a
Serway 32.38 (5 points)
(0.5 point)
The rate of energy being delivered (power) by battery is simply given by
Pbat = V I = 3 22 = 66.0 (W).
4.1.b
(0.5 point)
The power of resistor is
P = I 2 R = 32
Ph1c Practical Homework Solutions 1
Spring 2014
1.1
1.1.a
Serway 30.5 (5 points)
(2.5 points)
You can work the problem out from integrating the Biot-Savart law or by simply taking the result from
example 30.1 (equation 30.4). The magnetic eld due to one w
Ph1c Practical Homework Solutions 6
Spring 2014
6.1
6.1.a
Serway 33.48 (5 points)
(2 points)
A step-down transformer reduces the voltage from input to output. For our case the number of turns on the
input side is 13 times the number of turns on the output
Ph1c Practical Homework Solutions 2
Spring 2014
2.1
Serway 31.14 (5 points)
The rst step is to calculate the magnetic ux inside of the solenoid. From Amperes Law, the magnetic
eld inside a solenoid is
B = 0 nI
where n is the number of turns per unit lengt
Physics 1c practical, 2015
Homework 5 Solutions
Serway 33.42
L = 20.0 mH, C = 107 F , R = 20, Vmax = 100V
a
The resonant frequency for a series RLC circuit is f =
1
2
1
LC
= 3.56 kHz
b
At resonance Imax =
Vmax
R
=5A
c
Q=
0 L
R
= 22.4
d
VL,max = XL Imax =
Homework for ee/cs/est 135
Steven H. Low
EAS, Caltech
January 22, 2013
1
Transformers (Chapter 5, BV)
From text Bergen and Vittal, 2nd Ed., Chapter 5
5.2, 5.3, 5.4, 5.5, 5.6, 5.10, 5.12
D-Y per-phase equivalent circuit. The per-phase equivalent circuit o