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 poin
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 lectu
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
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Bounding complex
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
Welco
ACM 95/100a
Week 2 lecture notes
Dan Meiron
Caltech
January 15, 2017
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ACM 95/100a - Methods of Applied Mathematics for the Physical
January
Sciences
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Branch points
In
ACM 95/100a
Week 4 lecture notes
Dan Meiron
Caltech
January 26, 2017
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January
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Using the genera
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
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
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
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
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
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 =
Z
June 8, 2015
PHYSICS 1c Prac FINAL EXAM
This is an OPEN BOOK exam with the following limitations: You may consult only
your textbook (Serway & Jewett), note
Physics 1c Practical, Spring 2015
Quiz 4 Solutions
June 2, 2015
1
(a)
(10 points)
(2 points)
It will first be useful to calculate the Lorentz factors
1
2
Now, since L0 = L, we find that:
=q
=q
1
v12
c
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
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
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
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
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+
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 o
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
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 = 2.9
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
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
exam
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
i
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
Physics 1c Practical, Spring 2015
Homework 8 Solutions
May, 2015
39.6 (5 points)
(a) (2.5 points)
L = L0
r
1
p
v2
=
1
1
c2
0.92 m = 0.436 m
(1)
(b) (2.5 points)
Becomes longer, but still shorter than
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 Im
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