Download 360K 1200J 900J

Physics 2020
Spring 2008
Stephan LeBohec
1
FINAL
Name:_____________________________________
TA (circle one): Aaron
Akiko
Jacque
Student ID #:___________________________
Matt
A.
There are 1.2 moles of a monatomic ideal gas is at an initial temperature of
heat is added to the gas and 900J of work is done on the gas.
360K . Then 1200J of
A.
1.
[6 points] What is the change in internal energy during this process?
Internal energy change:  U =Q−W =1200J −−900J =2100J
A.
2.
[7 points] What is the final temperature of the gas?
2U
2×2100J
3
=
=140K
 U = n R  T so  T =
3 n R 3×1.2×8.31 J⋅K −1⋅mol−1
2
T Final=T Initial  T =360140=500K
B.
A thermal engine produces 2000J of work while consuming
released in the air at a temperature of 300K .
B.
B.
10000J of heat with any excess heat
1.
[7 points] How much heat is released in the air?
Q H =W Q C so QC =Q H −W =10000J−2000J=8000J of heat released to the cold reservoir
2.
[7 points] What is the efficiency of the engine?
Efficiency e=
W
2000J
=
=0.2 or 20%
Q H 10000J
B.
3.
[8 points] For a Carnot engine with the same efficiency and cold reservoir temperature, what is
the temperature of the hot reservoir?
Carnot efficiency e Carnot =1−
TC
Tc
300K
=375K
so T H =
with T C =300K and e=0.2 , T H =
1−0.2
TH
1−eCarnot
Physics 2020
Spring 2008
Stephan LeBohec
2
FINAL
Name:_____________________________________
TA (circle one): Aaron
Akiko
Student ID #:___________________________
Jacque
Matt
C.
[8 points] A famous painter with a bandage on his left ear is
contemplating his human condition in a pair of plane vertical mirrors. The mirrors
°
are making a 90 angle as in the figure which is a top view of the scene. The
triangle represents the painter's nose and the cross represents the bandage. You are
an artist too! On the figure, draw the images of the painter including nose and
bandage. How many images of the painter are there? How many images of the
artist have the bandage on the left ear?
There are 3 images, only one of them has the bandage on the left side like the
original.
D.
A telescope has an objective lens with a diameter 8cm and a 1m focal length. This telescope is used
to observe the Sun from Titan, one of the moons of Saturn. Titan is 1.4×1012 m from the Sun. The Sun has a diameter
of 1.4×109 m and a luminosity of 3.8×1026 W .
D.
1.
[7 points] What is the intensity of the radiation of the Sun at the distance of Titan?
2
26
12
2
−2
I =P / 4 r =3.8×10 W / 4 1.4×10 m =15.4 W⋅m
D.
2
[7 points] What is the diameter of the image of the Sun formed by the objective lens?
9
hi
di
ho
1.4×10 m
−3
=−
=10 m≡1mm is the size of the image.
and so hi =−d i =−1m ×
12
ho
do
do
1.4×10 m
D.
3.
[7 points] What is the angular diameter of the sun as seen through the telescope if the eyepiece
has a 10mm focal length?
' =
9
fo
1m
−1 1.4×10 m
°
=
tan 
=5.73 is the angular diameter of the sun as seen through the telescope.
12
fe
0.01m
1.4×10 m
D.
4.
[8 points] What is the size in kilometers of the smallest detail of the solar surface that can be
observed from Titan using this telescope and observing a wavelength =469nm (Hint: the limitation comes from
diffraction through the circular objective lens)?
The diffraction limited resolution angle in radians is
2×1.22
1.22

so the smallest detail that can be seen on the sun is
D
−9

469×10 m
12
7
d 0=2.44×
×1.4×10 m=2×10 m≡20,000 km
D
0.08
Physics 2020
Spring 2008
Stephan LeBohec
Name:_____________________________________
TA (circle one): Aaron
E.
1.
3
FINAL
Akiko
Jacque
Student ID #:___________________________
Matt
[8 points] What is the amount of work required to bring a charge
q ' =0.5×10−6 C from infinity to point M? ( q=2×10−6 C and d =0.2m )
W =q ' V f −V i  and we bring q ' from infinity so V i =0 and
qq ' 3qq ' kqq '
kqq '
W =k 
−
=
1−3/ 2=−
d
2d
d
2d
9
−6
−6
8.99×10 ×2×10 ×0.5×10
W =−
=−0.0225J
2×0.2
E.
2.
[10 points] What is the magnitude of the electric force on charge
q ' =0.5×10 C at point M? ( q=2×10−6 C and d =0.2m )
−6

2
2


 ∣=∣q '∣⋅∣ E
 ∣ and ∣E
 ∣= E 2x E 2y =  kq   3kq  = kq 1 9 = kq 25 = 5kq
∣F
16 d 2 16 4d 2
d2
4d 2
d2
9
−6
−6
 ∣= 5kqq2 ' = 5×8.99×10 ×2×102 ×0.5×10 =0.281N
∣F
3d
4×0.2
F.
schematic is
1.
[6 points] If the resistance measured between A and B on the
75  , what is the resistance value R?
1
R AB=R  R75=75 and so R=75/ 3=25 
2
F.
2.
[6 points] A battery is now connected between A and B and there is a 0.20A current flowing
through the resistance R shown in the dashed rectangle. What is electric potential difference between points B and C?
There are 0.100A flowing in each leg and V BC =0.1A×75 =7.5V
F.
3.
[6 points] A battery is still connected between A and B and there is a 0.20A current flowing
through the resistance R shown in the dashed rectangle. What is the electric potential difference between points C and D?
Equivalently, V BD =5V and consequently V CD=V BC −V BD=0V
Physics 2020
Spring 2008
Stephan LeBohec
4
FINAL
Name:_____________________________________
TA (circle one): Aaron
Akiko
Jacque
Student ID #:___________________________
Matt
G.
In the mass spectrometer represented on the figure, particles
with charge q=2e and mass m=4u ( 1u=1.66×10−27 kg ) are
accelerated by an electric potential difference of 1200V . Once through the
accelerating region, the particles enter a region where they are deflected
toward a fluorescent screen by a uniform magnetic field of magnitude
∣
B∣=0.16T . The impact point on the screen is at a distance d from the
location where the beam entered the magnetic field region.
G.
1.
[4 points] For the described mass spectrometer to
work should the electric field 
E in the accelerating region be pointing up or
down on the figure? (circle one)
↑
↓
G.
2.
[6 points] For the described mass spectrometer to work should the magnetic field 
B in the
magnetic deflecting region be pointing in or out of the figure? (circle one)
⊗
G.
⊙
3. [10 points] What is the speed of the particles when entering the magnetic field region?
1
2
The kinetic energy K =qV = mv
2
so v=
 
2qV
2×2×1.6×10−19×1200
5
−1
=
=3.4×10 m⋅s
−27
m
4×1.66×10
G.
4. [5 points] What is the speed of the particles when they hit the screen?
The magnetic force does not work on a free particle so the speed remains unchanged,
G.
v Screen=3.4×105 m⋅s −1
5. [10 points] What is the distance d?
m⋅v 4×1.66×10−27 ×3.4×105
=
=0.044m . The
The magnetic force is q⋅v⋅B=m⋅v / r so the giro-radius is r =
−19
q⋅B
2×1.6×10 ×0.16
distance d is twice the giro-radius so d =0.088m ≡8.8cm
2
Physics 2020
Spring 2008
Stephan LeBohec
FINAL
Name:_____________________________________
TA (circle one): Aaron
Akiko
Jacque
5
Student ID #:___________________________
Matt
H.
Three identical light bulbs with a resistance R are used in the three circuits A, B and C in the figure. The
internal resistance of the batteries can be neglected.
H.
1.
[6 points] How should one set V A and V B so that the light bulbs in circuit A and B shine
with the same brightness?
 V A=4V B
 V A=2V B
 V A=  2 V B
 V B =V A
 V B = 2V A
 V B =2V A
 V B =4V A
In circuit A the voltage across the light bulb is V A while in circuit B it is 2V B so V A=2V B
H.
2.
[6 points] How should one set V A and V C so that the light bulbs in circuit A and C shine
with the same brightness?
 V A=4V C
 V A=2V C
 V A=  2 V C
 V C =V A
 V C = 2 V A
 V C =2V A
 V C =4V A
V
In circuit A the voltage across the light bulb is
and in circuit C it is V C so V A=V C
A
H.
3.
[6 points] Assuming V A=V B =V C rank circuits A, B and C from the dimmest to the
brightest, using an equal sign to indicate light bulbs shining with the same brightness.
 A<B<C
 A<C<B
 B<C<A
 B<A<C
 C<A<B
 C<B<A
 A=B=C
 A=B<C
 A=C<B
 B=C<A
 A<B=C
 B<A=C
 C<A=B
In circuit A and C the light bulb is under the same voltage while in B the voltage is twice as large so A=C<B
H.
4.
[6 points] Assuming the electromotive force V A=V B =V C are all provided by identical
batteries with a fixed amount of charge available in their life time, indicate the order in which circuits A,B and C will stop
working,using an equal sign to indicate circuits that will stop working at the same time.
 A<B<C
 A<C<B
 B<C<A
 B<A<C
 C<A<B
 C<B<A
 A=B=C
 A=B<C
 A=C<B
 B=C<A
 A<B=C
 B<A=C
 C<A=B
Current flowing through the batteries in B is twice as much as in A where it is twice as much as in C so B stops first, then
A and then C: B<A<C
Physics 2020
Spring 2008
Stephan LeBohec
6
FINAL
Name:_____________________________________
TA (circle one): Aaron
Akiko
Student ID #:___________________________
Jacque
Matt
I.
We consider a solenoid with n=7958 loops per meter of
length. The current I S driven through the solenoid changes as indicated on
the graph. A coil with N =2000 circular loops of radius r =0.0488m is
wrapped around the solenoid. The coil has a resistance of R=12 .
I.
with time.
1.
[8 points] Graph the evolution of the magnetic field
Magnetic field in a solenoid B=0⋅n⋅I . The current reaches 4A and so
the magnetic field is going to reach 4 ×10−7×7958×4=0.04T
I.
2.
[8 points] Graph the evolution of the magnetic flux through the coil.
= A⋅B⋅cos = r 2 B⋅cos  but here =0 so =r 2 B the
magnetic field reaches 0.04T so the flux reaches
× 0.04882 ×0.04=3×10−4 T⋅m 2
I.
3.
[9 points] As the current I S in the solenoid changes, the induced current I I changes too.
Complete table by writing “same” when I I flows around the coil in the same direction I S flows around the solenoid,
“opposite” when I I flows around the coil in the opposite direction I S flows around the solenoid or “null” when
I I =0 .
I.
4.
IS
increases
is constant
decreases
II
opposite
null
same
[10 points] Graph the magnitude of the induced current
=0 to =3×10−4 T⋅m 2 in

3×10−4
∣=2000
=2.4V and the
 t=0.025s so ∣EMF∣= N∣
t
0.25
EMF 2.4V
=
=0.2A . When the flux decreases, the change
current is I =
R
12 
When the flux increases it goes from
in flux is the same but it takes twice as long and the current is twice smaller.
II .