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9:; 1. Reverse Engineering. What is the value of the following following mathematical operation operation :
1.23 − 45. 45.6/789 + 10 × 11. 11.12/ 12/131415 if the value is rounded to the correct number of significant digits? A. 1
C. 1.17
B. 1.2
D. 1.173
E. 1.17228
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45.¯6 45. 11..1¯2 11 ¯ 1.23 − ¯ + 10 × 789 12141¯6
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(2)
=
(3 )
1.17
/>57"%&52 C352#$%$ 3. INC Circle. In your projectile motion experiment, you obtain an equation for the vertical + bx − cx2 , where a, a, b, c are constants. (y ) vs horizontal (x) position of the form: y = a + bx Which of the following is a linear graph? A. y vs x
C. y vs a − cx2
B. y vs x2
D. y vs a + a + bx bx
E. y vs (x (x −
b 2 2c
)
/>57"%&52 C352#$%$ •
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b
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+ a b2
b
x − x + 2 c 4c
= −c x −
•
(1) (2)
b
2c
(3)
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/>57"%&52 C352#$%$ 2nd Polate. Polate. 4. 2nd Galaxies Galaxies expanding expanding with with recess recession ion veloc velocity ity v at appar appar-ent distance distance r obeys the Hubble Law: v = H r where H is the Hubble constant. stant. For For individ individual ual galaxies galaxies in the unive universe rse,, the exper experimen imental tal plot plot of r vs v is shown shown at the right right.. How How far is a galaxy in the constellation Virgo if it has a recession velocity of 1200 km/s?
A. 4.8 × 1023 [m]
C. 4.0 × 1017 [m]
B. 1.0 × 1022 [m]
D. 3.0 × 10
−15
[m]
−18
E. 2.5 × 10
[m]
/>57"%&52 C352#$%$ •
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y [×1023 m]
= 4.0[×1023−6 s−1 ] · x · x[[×106 m/s] m/s] + 0. 0 .1[×1023 m]
(1)
y [×1023 m]
= 4.0[×1023−6 s−1 ] · (1 · (1..2)[×106 m/s] m/s] + 0.1[×1023 m] = 4.8 × 1023 m + 0. 0 .1[×1023 m]
(2) (3)
9,5$0>%3E ;,-%&,$ 21. Least Count. What is the least count of the Vernier Vernier Caliper? Caliper? A. 0.20 [mm]
C. 0.05 [mm]
B. 0.10 [mm]
D. 0.02 [mm]
E. 0.01 [mm]
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The Vernier caliper, as shown in Figure 3, has a main scale and a Vernier scale (fractional scale). The main scale has a unit mark of 1. 1 .00 mm which which is furthe furtherr divide divided d into into 50 divisio divisions ns in the Vernie Vernierr scale. The least count of the Vernier caliper is therefore 1. 1.00mm/ 00mm/50 = 0. 0.02 mm with each each unit mark in the Vernier scale equal to the least count. Since the uncertainty of a device is half its least count, the absolute uncertainty of the Vernier caliper is 0. 0.02mm/ 02mm/2 = 0.01 mm. mm.
:CR9 29. Position. Which Which of the following following lines/curv lines/curves es as seen from the Labquest Labquest best depicts the shape of the position vs. time plot as the ball slides down the ramp? A.
B.
C.
D.
E.
30. Velocity. Which of the following lines/curves as seen from the Labquest best depicts the shape of the velocity vs. time plot as the ball slides down the ramp? A.
B.
C.
D.
E.
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!>?\,&D2, 9?D?3 [m] at an angle of 35 35.00 9. UP. UP. A ball was launched from a ramp with initial height Y 0 = 0.30 [m]
◦
with with an init initia iall velo veloci city ty of abou aboutt 3.00
hmi s
and and land landss on a flat flat sur surface ace whic which h is 0.25[m] away.
What is the elevation of the surface where the ball has landed? A. 0.31 [m]
C. 0.45 [m]
B. 0.38 [m]
D. 0.62 [m]
E. 0.76 [m]
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.
The path of the projectile can be determined by eliminating the time component from these equav0 as ~ v0 = v 0 ˆi + v0 jˆ, we can show that tions. Decomposing the initial velocity ~ g y = y 0 + x tan θ − 2 x2 , (8) 2v0 cos2 θ x
y
where θ = tan−1 (v0 /v0 ). The path of the projectile is parabolic opening downward [3]. y
•
x
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!>?\,&D2, 9?D?3 10. Projectile range. A ball was launched from a ramp with an initial height at an angle of 35 35 .00 with an initial velocity of about 3 .00 ◦
hmi s
Y 0
= 0.30 [m] [m]
and lands on a flat surface
which is elevated to 0.70[m]. How far did the ball travel (range)? A. 1.2 [m]
C. 0.21 [m]
B. 0.40 [m]
D. 0.12 [m]
E. 0.04 [m]
!>?\,&D2, 9?D?3 •
•
•
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x
.
.
Y
When a projectile is launched from a level surface, the final (vertical) position is the same as the initial position (y (y = y = y 0 ). For this case, the range of the projectile R is: R =
sin 2θ v02 sin . g
(9)
4?031 13. Beat 13. Beat frequency. If you simultaneously strike two tuning forks having frequencies of 305 305 and 315 [Hz] [Hz] respectively, what would be the beat frequency? A. 620 [Hz]
C. 20. [Hz]
B. 615 [Hz]
D. 5.0 [Hz]
E. 10. [Hz]
4?031 •
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O L I P
f beat beat = | f 1 − f 2 | ,
(14)
which can be experimentally determined by measuring the period of the beat pattern and using the definition f beat beat =
1 T beat beat
.
(15)
Humans Humans can detect up to about 15 to 20 beats p er second only [4]. Beyond Beyond this, the fluctuations fluctuations in loudness would be too rapid to be distinguished. •
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◦
25. Wherelalooo. Where is the resultant force due to String A and String B located? A. 33
C. 48
B. 42
D. 57
◦
◦
◦
◦
E. 71
◦
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res
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=
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(1) (2) (3) (4)
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(5)
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(6)
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46
◦
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