Physics 221 2007S
Exam 1 Solutions
[1]If a 200kg box is subject to a net force of magnitude 100N, what is the magnitude of
its acceleration?
(A) 0.25 m/s 2
(B) 0.50 m/s 2
(C) 1.0 m/s 2
Answer (B): Using Newton’s second law a =
(D) 2.0 m/s 2
(E) 4.0 m/s 2
F 100N
m
=
= 0.5 2
m 200kg
s
[2] A car is traveling a circular track at a constant speed of 20m/s. If the magnitude of the
acceleration of the car is 1 m/s 2 , what is the radius of the track?
(A) 50m
(B) 100m
(C) 200m
(D) 400m
Answer (D): Centripetal acceleration is given by a =
Page 1 of 21
(E) 800m
v2
v 2 (20 ms ) 2
so R = =
= 400m .
R
a
(1 sm2 )
Physics 221 2007S
Exam 1 Solutions
[3] An elevator has a mass of 600kg, not including passengers. The elevator is designed
to ascend, at a constant speed, a vertical distance of 20.0m (i.e. five floors) in 16.0s, and
is driven by a motor that can provide up to 30kW of power. What is the maximum
number of passengers that can ride in the elevator? Assume that an average passenger has
a mass of 65kg.
(A) 28
(B) 24
(C) 20
(D) 16
(E) 12
Answer (A): The force that the motor must exert on the elevator is equal to the weight of
the elevator since it is moving at a constant speed. Since P = Fv the force exerted by the
motor and hence the weight of the elevator is
W = F = P / v = (30000W ) /(20m /16 s ) = 24000 N
The mass of the elevator is thus m = W / g = (24000N) /(9.8 sm2 ) = 2449kg
From the problem, the number of passengers n is
n = (m − 600kg) / 65kg = (2449kg − 600kg) / 65kg=28.4
The limit is thus 28 passengers.
[4] A person pulls a 20-kg block up a ramp of height h = 10 m and angle θ = 20° at
constant speed. The rope makes an angle φ = 35° with the ramp as shown and has a
tension of 160N . Find the work done by kinetic friction.
(A) −661J
(D) −1880J
(B) −740J
(E) −5800J
(C) −980J
Answer (D):
The length of the ramp is
h
10m
Δx =
=
= 29.3m
sin θ sin 20o
The work done by the tension is
WT = T Δx cos ϕ = (160N)(29.3m) cos(35o ) = 3840J
The work done by gravity is
Wg = − mgh = −(20kg)(9.8 sm2 )(10m) = −1960J .
The normal force does no work since it is
perpendicular to the ramp. By the work energy theorem, the total work on the block is
zero because its kinetic energy does not change as the block is moving at a constant
speed. Thus
Wnet = 0 = Wg + WT + Wk
∴Wk = −(WT + Wg ) = −((+3840J) + (−1960J)) = −1880J
Page 2 of 21
Physics 221 2007S
Exam 1 Solutions
[5] A bullet of mass 1g is fired at a block of wood traveling at a speed of 300 m/s. It
comes to rest 60μs after hitting the surface of the wood. If you assume that the wood
slows down the bullet exerting a constant force against the bullet’s motion, what is the
magnitude of that force?
(A) 2500N
(B) 5000N
(C) 7500N
(D) 10000N
(E) 25000N
Answer (B): The acceleration of the bullet is (change in velocity)/(time interval) so
m
m
a = (300 ) /(60μ s ) = 5.0 ×106 2 in the direction against the motion of the bullet. From
s
s
m
Newton’s second law, the force on the bullet is F = ma = (10−3 kg)(5.0 ×106 2 ) = 5000N
s
[6]A lady bug and a gentleman bug are sitting on a circular turntable with radius 50cm
which is rotating clockwise at a constant angular velocity. The ladybug is at the rim of
the turntable while the gentleman bug is half way between the center and the rim. If the
lady bug is moving at a speed of 0.10m/s, what is the magnitude of the angular velocity
of the gentleman bug?
(A) 0.05 rad/s
(B) 0.10 rad/s
(C) 0.20 rad/s
(D) 0.30 rad/s
(E) 0.40 rad/s
Answer (C): The angular velocity of the gentleman bug is the same as that of the lady
bug because they are on the same turn table. This angular velocity is
ω = v / R = (0.10 ms ) /(0.50m) = 0.20rad/s
Page 3 of 21
Physics 221 2007S
Exam 1 Solutions
[7]Sally runs 1 km north at 6 m/s and then runs 2 km south at 4 m/s. What is the
magnitude of Sally’s average velocity for this trip?
(A) 0.0 m/s
(B) 1.5 m/s
(C) 2.0 m/s
(D) 3.0 m/s
(E) 5.0 m/s
Answer (B): The total displacement of Sally’s trip is 1km south. The first part of her trip
takes an amount of time t1 = 1000m /(6m/s) = 167s . The second part of her trip takes an
amount of time t2 = 2000m /(4m/s) = 500s . The total time is thus ttot = t1 + t2 = 667s . The
magnitude of the average velocity is thus (1000m)/(667m/s)=1.5m/s.
[8]An object’s height as a function of time is y (t ) = (120 m) + (80 m/s)t − (20 m/s 2 )t 2 .
What is the highest point this object reaches?
(A) 100m
(B) 150m
(E) None of the above
(C) 300m
(D) 400m
Answer (E): The extremum is given by y’=0. Taking the derivative,
y ' = (80 ms ) − (40 sm2 )t
solving for y ' = 0
t = (80 ms ) /(40 sm2 ) = 2s
This is in fact a maximum because y '' = −40 sm2 is negative. Plugging this time back into
the expression for y we get the height at the maximum:
ymax = (120 m) + (80 m/s)t − (20 m/s 2 )t 2
= (120 m) + (80 m/s)(2s) − (20 m/s 2 )(2s) 2
= 200m
(Which is not one of the choices A-D)
Page 4 of 21
Physics 221 2007S
Exam 1 Solutions
[9]A 100kg person stands on top of a box that is inside the cabin of an elevator. The
elevator has an upward acceleration of 3 m/s 2 . What is the magnitude of the net force
acting on the person?
(A) 100N
(B) 300N
(C) 680N
(D) 980N
(E) 1280N
Answer (B): By Newton’s second law Fnet = ma = (100kg )(3 sm2 ) = 300 N .
[10] The figure shows the overhead view of two stones that travel in circles over a
frictionless horizontal surface. Each stone is tied to a cord whose opposite end is
anchored at the center of the circle. The radius of the circle of stone B is 2 times the
radius of the circle of stone A. The tension in the long cord is 2 times the tension in the
short cord. The period of stone B is 4 times the period of stone A. What is the ratio
(mass of stone B):(mass of stone A)
(A) 16:1
(D) 1:4
(B) 4:1
(E) 1:16
(C) 1:1
Answer (A): The centripetal force is F = mRω 2 = 4π 2
Thus
B
A
mB FB TB2 RA
2⎛1⎞
=
= 2 ⋅ ( 4 ) ⎜ ⎟ = 16
2
mA FA TA RB
⎝2⎠
Page 5 of 21
mR
T2
∴m =
1 FT 2
4π 2 R
Physics 221 2007S
Exam 1 Solutions
[11]A rubber ball is tossed horizontally from a window 5m above a parking lot. When it
strikes the ground it rebounds. Which of the following graphs best represents the ycomponent of the ball’s velocity as a function of time? The +y direction is up and neglect
air resistance.
vy A
vy
vy
B
C
Time
Time
vy
vy
D
Time
Time
E
Time
Answer (C): While the ball is in freefall, before and after the bounce, it is accelerating
downwards so the velocity graph should have a constant negative slope. During the
bounce the velocity will change rapidly from downwards to upwards. Only graph (C) has
these features.
Page 6 of 21
Physics 221 2007S
Exam 1 Solutions
[12]Robin Hood fires an arrow at velocity v0=30m/s at an angle of θ=30o above the
horizontal from the top of a cliff of height 20m. The sheriff of Nottingham stands at
distance d from the base of the cliff. What is the speed of the arrow when it strikes the
sheriff? Neglect air resistance.
(A) v=36 m/s
(B) v=42 m/s
(C) v=46 m/s
(E) Cannot determine the speed without knowing d.
Robin
Hood
v0=30m/s
(D) v=50 m/s
θ=30o
y
h20m
v=?
d
Sheriff of
Nottingham
x
Answer (A): Using the v-squared equation in three dimensions:
G G
v 2 − v02 = 2a ⋅ Δr
m
G
G
Taking the x-axis horizontal and y-axis vertical, a = (0, −9.8 2 ) and Δr = (d , −20m) so
s
G
G
v = v02 + 2a ⋅ Δr = (30 ms ) 2 + 2(−9.8 sm2 )(−20m) = 36 ms
Note that the distance d is determined by the angle θ but to get the speed of impact we
don’t have to know either of these quantities.
Page 7 of 21
Physics 221 2007S
Exam 1 Solutions
[13]
G
G
Vectors A and B start at a corner of a
G
cube of edge length 1. Vector A lies
G
along the edge of the cube. Vector B
goes to the opposite corner of the cube.
G
G
What is the angle between A and B ?
(A) 45°
(B) 55°
1
(C) 60°
(D) 70°
(E) 90°
Answer (B): Let us take the origin to
be the starting part of the vectors in the
picture
and the x axis parallel to the vector A. In this coordinate system the vectors are
G
A = (1, 0, 0)
G
B = (1,1,1)
thus the cosine of the angle between the vectors is given by:
G G
A⋅ B
(1, 0, 0) ⋅ (1,1,1)
1
cos θ =
=
=
AB | (1, 0, 0) || (1,1,1) |
3
1
∴θ = arccos
= 55o
3
[14]
A particle of mass m = 10.0 kg moves along the x-axis with time dependent acceleration
ax = ( 3.00 m/s 4 ) t 2 . The velocity of the particle at t = 0 is -3.00 m/s in the +x direction.
What is the velocity of the particle at t = 2.00s?
(A) 0 m/s
(B) 5 m/s
(C) 8 m/s
(D) 9 m/s
Answer (B): Acceleration is the derivative of velocity thus:
t final
v = v0 +
∫
a dt
0
2.00 s
= −3.00 + (3 )
m
s
m
s4
∫
t 2 dt
0
2s
⎡t3 ⎤
= −3.00 + (3 ) ⎢ ⎥ = −3.00 ms + 8.00 ms
⎣ 3 ⎦0
= 5 ms
m
s
m
s4
Page 8 of 21
(E) 21 m/s
Physics 221 2007S
Exam 1 Solutions
[15] Two cars move along parallel tracks on a straight road. Shown below are the
snapshots taken every second. Assume that the motion between snapshots is what one
would assume from the image, nothing strange. Which of the following best describe the
direction of the velocity of the rectangular car relative to the square car at t = 2 s and at t
= 3 s?
(A) ← , →
(B) ← , ←
1s
(C) → , →
2s
1s
(D) ← , 0
3s
2s
3s
(E) → , 0
4s
4s
Answer (B): At 2 and 3 seconds it is clear that square car is faster than the rectangular car
and both moving to the right. Since we want the velocity of the rectangular car relative
to the square car, using eqn. 3.36 from the text:
G
G
G
vrec / ground = vrec / squ + vsqu / ground
G
G
G
∴ vrec / squ = vrec / ground − vsqu / ground
G
G
vrec / ground + (−vsqu / ground )
Drawing the vectors (the same diagram applies to both t=2s and t=3s):
G
vrec / squ
=
G
vrec / ground
+
G
(−vsqu / ground )
Therefore both arrows are to the left and so the answer must be (B)
Page 9 of 21
Physics 221 2007S
Exam 1 Solutions
The following information applies to questions [16] and [17]: In the figures below,
the force applied in both cases on the system has the same magnitude and is horizontal.
Block A has twice the mass of block B. There is no friction between the blocks and the
surface underneath them.
F
mA
mA
mB
mB
Figure 2
Figure 1
[16]What is the magnitude of the net force on block A in figure 1 and figure 2,
respectively?
Answer (D): Consider first Figure 1. The acceleration of
the two blocks is the same as the acceleration of the
F
(A) F ,
whole system. Using Newton’s 2nd law on the whole
3
system , one can determine the acceleration. This
2F
F
(B)
,
acceleration is a = F /(3mB ) . Using Newtons 2nd law on
3
3
F
2F
2F
F
=
. In Figure 2 the
block A, FA = mA a = 2mB
(C)
,
3
m
3
B
3
3
magnitude of the net force on the whole system is the
2F
2F
(D)
,
same as Figure 1 so the above argument applies and the
3
3
net force on A is the same.
F
F
(E)
,
3
3
[17] For the same figures in the previous problem, which of the following is TRUE?
(please read carefully)
(A) The force on A by B is smaller in figure 1 than in figure 2.
(B) The force on A by B is the same in both figures.
(C) In figure 1, the force on A by B is smaller than the force on B by A.
(D) In figure 2, the force on A by B is smaller than the force on B by A.
(E) None of the above.
Answer (A): The force between the blocks in figure 1 is FAB1 = mB a since this is the only
horizontal force on block B. In figure 2 the force between the blocks is
FAB 2 = mA a = 2mB a since this is the only horizontal force on A. Thus option (A) is true
hence (B) is false. (C) and (D) are false due to Newton’s 3rd law because this law implies
FAB = FBA no matter what.
Page 10 of 21
F
Physics 221 2007S
Exam 1 Solutions
[18] A block of mass 2kg moving along the x-axis is subject to a position dependent force
given by
G
F = Ax 2 iˆ
where A = 2 kg/(ms 2 ) . Initially the block is at x=0m and has a velocity of (10 m/s) iˆ .
When the block reaches x=4m, what is its kinetic energy?
(A) 43J
(B) 57J
(C) 100J
(D) 143J
(E) 228J
Answer (D): By the work energy theorem, the change in kinetic energy is equal to the
work done on the block. In this case,
K f = Ki + W =
1 2
mv0 +
2
4m
∫ Ax
2
dx
0
4m
1
= (2kg)(10 ms ) 2 + (2 mskg2 ) ⎡⎣ 13 x 3 ⎤⎦
0
2
= 143J
Page 11 of 21
Physics 221 2007S
Exam 1 Solutions
[19] A 10-kg weight hangs from two ideal massless ropes as shown below. Find the
tension on the left rope.
(A) 33 N
(B) 49 N
(C) 85 N
(D) 98 N
(E) 113 N
Answer (B): This question may be simplified considerably by noting that the two strings
intersect at a right angle. If we take the coordinate system shown the y-component of the
weight is mg sin θ where θ=30° as shown. Applying Newton’s 2nd law in the y-direction
we find that therefore the net force in the y direction is Fnet y = T − mg sin θ therefore
T − mg sin θ = ma y = 0
because the 10kg object is not accelerating. Thus
T = mg sin θ = (10kg)(9.8 sm2 ) sin 30o = 49N
This problem can also be done using the coordinate system with the x-axis horizontal.
y
30°
=θ
60°
x
T=?
mg cos θ
mg sin θ
10kg
mg
Page 12 of 21
Physics 221 2007S
Exam 1 Solutions
[20] City B is 1000km east of city A. A light plane can fly at a speed of 400km/hr with
respect to the air. If there is a constant wind at a speed of 300km/hr blowing in a direction
30° north of east, what is the time it takes for the plane to make the trip from A to B?
(A) 2.65 hr
(B) 2.20 hr
(E) The trip as described is not possible.
(C) 1.59 hr
(D) 1.46 hr
Answer (C): The velocity of the plane with respect to the ground by
G
G
G
v plane, ground = v plane ,air + vair , ground . This relationship is shown in the figure below. We know
G
G
the magnitude and direction of vair , ground and just the magnitude of v plane,air . From the
G
diagram, we see that we can think of the resultant vector v plane, ground as being composed of
G G
two pieces, P , Q .
The height of the triangle is vair , ground sin θ so
P = vair , ground cos θ
Q = v 2plain ,air − (vair , ground sin θ ) 2
v plane, ground = P + Q
= vair , ground cos θ + v 2plain ,air − (vair , ground sin θ ) 2
o
o 2
km 2
km
= (300 km
hr ) cos 30 + (400 hr ) − (300 hr sin 30 )
km
km
= 260 km
hr + 371 hr = 631 hr
The time of the trip is thus 1000km/(631 km
hr )=1.59hr . You can also solve this triangle
using law of cosines, expressing the vectors in components or other methods
G
vair , ground
G
v plane,air
300km/hr
vair , ground sin θ
θ=30°
G
P 260km/hr
G
v plane, ground
631km/hr
Page 13 of 21
400km/hr
G
Q 371km/hr
Physics 221 2007S
Exam 1 Solutions
[21] For the vectors shown below, which of the following statements is TRUE?
G G G G
(A) A ⋅ C = A ⋅ D
G G G G
(B) A ⋅ B > A ⋅ C
G G G G
(C) C ⋅ B = C ⋅ D
G G
G G
(D) A ⋅ B = − A ⋅ D
G G
G G
(E) B ⋅ C = −C ⋅ B
A
D
B
C
Answer (D): Taking each square to be one unit, the vectors in component form are:
G
A = (6, 0)
G
B = (3,3)
G
C = (−3,3)
G
D = (3, 0)
G G
G G
Thus A ⋅ B = 18 and A ⋅ D = −18 hence equation (D) is true. The other equations are all
false.
Page 14 of 21
Physics 221 2007S
Exam 1 Solutions
[22] A block of mass 10kg lies at rest on an incline ramp tilted at an angle of 10º to the horizontal. The
coefficient of static friction between the block and the ramp is μS=0.4. The coefficient of kinetic friction
between the block and the ramp is μK=0.3. What is the magnitude of the frictional force on the block?
(A) 0N
(B) 17N
(C) 29N
(D) 39N
(E) 49N
Answer (B): First we need to see if the block slides or is held by static friction.
Assume first that the block is static. The free body diagram in this scenario is as shown
below where the block has 0 acceleration therefore the net force on it is 0.
Let us take the x-axis parallel to the ramp. From Newton’s 2nd law in the y direction, the
normal force is N = mg cos θ where θ is the angle of the ramp. The static friction limit is
fmax=μsN=μsmgcosθ.
Form Newton’s 2nd law in the x direction, the force of static friction must be
f s = mg sin θ because it must cancel the down-ramp component of the block’s weight.
This scenario is consistent only if:
f s < f max
⇔ mg sin θ < μ s mg cos θ
⇔ tan θ < μ s
⇔ tan10o = 0.18 < 0.4 ⇔ which is true
Thus, given the numbers in the problem this scenario is consistent. The force of static
friction is thus given as above: f s = mg sin θ = (10kg )(9.8 sm2 ) sin10o = 17 N
Note in particular that 39N which is the static friction limit is not the correct answer.
Newton’s 2nd Law
x-component:
mg sin θ − f s = 0
N
fs
mg sinθ
y-component:
N − mg cos θ = 0
y-axis
mg cosθ
mg
x-axis
θ=10°
Page 15 of 21
Physics 221 2007S
Exam 1 Solutions
[23]A 2kg cannon ball is fired from ground level at an angle of 45° with respect to the
horizontal. The ball hits a castle on a cliff which is 600m away horizontally and 200m
above the ground. What was the initial speed of the cannon ball? Neglect air resistance.
(A) 43 m/s
(B) 63 m/s
(C) 66 m/s
(D) 94 m/s
(E) 112 m/s
Answer (D):
Method 1: If there were no gravity, the cannon ball would pass over the castle at an
altitude of 600m which is 400m above the cliff. The time it takes to hit the castle must
therefore be the same as the time it takes for something to fall a distance of d=400m.
2d
Thus t =
= 9.0s . Horizontally, the cannon ball covers 600m in that time so the
g
horizontal speed is vx = (600m) /(9.0s) = 66.4 ms . The initial speed is thus
v = vx / cos(45o ) = 93.9 ms .
No gravity
trajectory
d=400m
Actual
trajectory
200m
θ=45°
600m
Page 16 of 21
Physics 221 2007S
Exam 1 Solutions
Method 2:
Let t be the time it takes to hit the castle. Taking the origin to be the point where the
cannon is fired, the x and the y components at time this time are:
x = vt cos θ
y = vt sin θ − 12 gt 2
The two unknown quantities are t and v but we have two equations so we can solve the
system. Starting by solving the first equation for t:
x
v cos θ
plugging this back into the second equation:
t=
y = x tan θ − 12 g
x2
v 2 cos 2 θ
gx 2
= x tan θ − y
2v 2 cos 2 θ
gx 2
∴ 2v 2 cos 2 θ =
x tan θ − y
∴
∴v =
=
=
gx 2
2 cos 2 θ ( x tan θ − y )
(9.8 sm2 )(600m) 2
2 cos 2 45o ((600m) tan 45o − 200m)
(9.8 sm2 )(600m) 2
(600m − 200m)
= 93.9
=
3.53 × 106
m3
s2
400m
m
s
Page 17 of 21
Physics 221 2007S
Exam 1 Solutions
G
G G
G G
G
[24] Let A and B be two vectors such that A + B = 2 and A − B = 4 . What is the value
G G
of A · B ?
(B) −1/ 2
(A) 0
(C) −1
Answer (E):
Expanding:
G G
G G
( A + B ) 2 = A2 + B 2 + 2 A ⋅ B
G G
G G
( A − B ) 2 = A2 + B 2 − 2 A ⋅ B
G G
G G
G G
Therefore ( A + B ) 2 − ( A − B) 2 = 4 A ⋅ B
Thus
G G
G G
G G
A ⋅ B = 14 [( A + B) 2 − ( A − B) 2 ]
= 14 [(2) 2 − (4) 2 ]
= −3
Page 18 of 21
(D) −2
(E) −3
Physics 221 2007S
Exam 1 Solutions
[25] Consider the system shown below made of three identical blocks of mass m, two
ideal, massless strings and an ideal, massless pulley. The friction between the blocks and
the table is negligible. A force of F=mg is applied to the end of the string as shown.
m
T1
m
T2
m
F=mg
Let T1 and T2 be the tension on each string, as shown in the figure. Which of the
following is true?
(B) 2T1 = T2 = mg
(C)2T1 = T2 = 23 mg
(A) T1 = T2 = mg
(E)T1 = T2 = 32 mg
(D)T1 = T2 = 23 mg
Answer (C): The net force on the three block system of mass 3m is mg to the right. The
acceleration is therefore a = g / 3 . The only horizontal force on the left block is T1 . This
block accelerates at the rate a so by Newton’s 2nd law this tension must be
T1 = ma = 13 mg . The only net horizontal force on the system consisting of the two left
blocks is T2 and the mass of this system is 2m so again, this force is
T2 = (2m)( 13 g ) = 23 mg . Thus (C) is the correct answer.
Page 19 of 21
Physics 221 2007S
Exam 1 Solutions
[26] Ball A is dropped from the top of a building of height h at the same instant that ball
B is thrown vertically upward from the ground. When the balls collide, they are moving
in opposite directions, and the speed of A is two times the speed of B. How far above the
ground does the collision occur?
(A) h/6
(B) h/3
(C) h/2
(D) 2h/3
(E) 5h/6
Answer (D):
Let t be the time after the balls are released when the collision occurs. Since the balls
collide:
y A (t ) = yB (t )
and from the condition on the velocities at collision:
v A (t ) = −2vB (t )
Using the constant acceleration equations we can expand the positions and velocities in
the above as follows:
v A = − gt
y A = h − 12 gt 2
vB = vB 0 − gt
yB = vB 0t − 12 gt 2
Plugging these into the above we obtain two equations in the two unknowns vB0 and t:
y A = yB ⇒ h − 12 gt 2 = vB 0t − 12 gt 2
⇒ h = vB 0 t
−v A = 2VB ⇒ gt = 2vB 0 − 2 gt ⇒ 3 gt = 2vB 0
Solving this system of two equations in the two unknowns we obtain
vB 0 =
h
t
∴ 3 gt = 2 ht
⇒ t2 =
2h
3g
plugging this into the equation for yA:
y A = h − 12 gt 2
= h − 12 g ( 32gh )
= 23 h
The height of the collision above the ground is therefore y = 23 h .
Page 20 of 21
Physics 221 2007S
Exam 1 Solutions
[27] [EXTRA CREDIT PROBLEM] A crate is dragged at constant speed along a
horizontal floor with a cable at an angle with the horizontal (see figure). Let these
symbols represent the magnitude of the following forces on the crate:
T = tension in the cable
f = friction force
N = normal force by the floor
W = weight of the crate
T
Which of the following is TRUE?
(A)
(B)
(C)
(D)
(E)
T>f
T<f
T<f
T>f
T>f
;
;
;
;
;
N<W
N=W
N<W
N=W
N>W
Answer (A): Consider the free body
diagram of the system. The block moves at
a constant speed so the net force on it is 0.
In the vertical direction, N+Ty=W so that
N<W. In the x direction
f k = Ty = T cos θ < T since cos θ is always
N
Ty T
fk
θ
less than 1.
Tx
W
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