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Unformatted text preview: Key I Page 1 (20 PTS) 1. In 1970, the year that the price of handheld calculators dropped below $500 each, there were
approximately 22.4 million slide rules in existence. (a) If slide rules were destroyed at the rate of ﬁfteen per minute, how many years would it take to
eliminate these 22.4 million slide rules? (7 W0 M QJ(’:om<:m—+><aa,> rm} w 5sitdt ml (b) If the shape of an aluminum slide rule can be accurately approximated by a box 4.00 mm thick,
30.0 mm wide, and 35.0 cm long, how many aluminum atoms are found in such a slide rule? Volume 1‘ (Li X (0'31“)(30xl0‘3mX35 XtO‘zm) = 4 "L x (05 m3
mass 2 (mass Mslt7):(vo\um) (2 .70 xto3k3m>(th(o“°m3): 01ml 9 NUMloer 0+ a’mms = W __._. 0 “34 J:
. m
(1550 atom (27x11?!) )1Molz
MoLL 0009 6 amounts“ Numwoi am; 2 2.52% x1024 ot’rws Page 2 (20 PTS) 2. Consider the time variation of velocity of an object moving along the x axis as in the graph
below. Graph the position versus time and the acceleration versus time for the object. Draw both
graphs. Assume that the object started 3.5 cm to the left of the origin. ’  W%)
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noII§===EIIIIIIIIIIIIIIIIIII Page 3 3. A ﬁsh swims along a straight, narrow river channel according to the equation
x= (14 00 m)— (9.50 m/s)t + (0. 250 m/s )3. (10 PTS) (a) Determine the average velocity for the ﬁsh’s travel from t = 10.0 s to t = 25.0 s. (4 PTS) (b) Determine the instantaneous velocity of the ﬁsh at t= 10.0 s and t = 25.0 s. Give both
values. (3 PTS) (c) Determine the average acceleration for the ﬁsh’s travel from t = 10.0 s to t= 25.0 s.  v9) _
“AVGIL V1 1M): : 0.50.": 1.! ft 135 V 1:5? 51 (3 PTS) (d) Determine the instantaneous acceleration of the ﬁsh at t = 10.0 s and t = 25.0 5. Give both
values. Page 4 4. At noon, a tiny island’s radar station locates a stationary ship at range 43.0 km and bearing 18
degrees clockwise from north. Also at noon, from the same radar station, an airplane ﬂying 4.90
km above sea level is located at range 32.0 km and bearing 210 degrees clockwise from north. (15 PTS) (a) Determine the number of kilometers separating the stationary ship and the ﬂying plane at noon.
 Wt 5:9!!! WW g am,
5' 1319"“ ‘lofiokm 43km 71° 0km
fl ~l6.oolm ~27.7 kw mm are Mo km 3‘: (Dawn)? «worm + (0 wt”
#l : (~1600ltyh)lt\ 4 (~17.7km)3 +(Mo km)k swam : 55/? = @q.1ﬁkm)€ +r(6%~cm)5 + (“ltowk m i‘Pu’K‘tOTShIE 4 5994(a‘hon if 5 ~A . _ l 1, 1.
mg («921 about the Swarm“ Wayward! = (21mm) +($%.6 lam) +(.M0km)
k direction ‘
3 SSS’Wﬂ let/Vt" ; i7q3 km! (5 PTS) (b) The airplane ﬂies with constant velocity (3.00 i + 4.00 j ~— 0.0200 R) m/s. Here, i is east, j is
north, and k is up. Determine the location of the airplane at 12:30 pm as measured from the
radar station. Please express this new position using unitvector (i, j, k) notation. A rim rum Vector 4 .13“, A4130: Allow + VAT
2?on : (’léOokm) f + 6217*”); + (4'70 [WU/L )j + (Loni—abomgﬁ 1 mm + (~31; )(som..)(lg.r,;; ) 37;)? +(+L;)(1~om;n)(/o_g:)(_62§ Page 5 5. A frog covers a horizontal distance of 1.15 m while Jumping to the ﬂat ground from a height of 25.0
cm above the ground. At the maximum height, which 15 61.5 cm above the ground, the frog catches a ﬂy. Before the frog lands, the ﬂy 1s swallowed.
(10 PTS) (a) Determine the time it takes the frog to make the Jump. VLﬁV=1QA
Xlta ji:25CW\ ﬂ 7y _'
O ‘ Vyl‘ = 7—( 7'” 51)(él'5cm 25”") X2 Y1 : 6115 cm V7; 2 i7J5Y35: : 2,67g “VS W: .
X3: [/5044 Y3 3 O cru/ Vyﬁ :Vyi +051
0: 267§M+C11L§Q1>AT V‘N'x‘ Vyil : 1 “V A) At 1 2_";’7%5 I»)? : 0.273 52
\I .L— z 7. "1351., 'O 0.6I'Sm ., ‘ ”in“ I .
7'1 ' O ( S )( ) ANSWQK iii?) Atm“ — 01735108541 : than: V = 4A1 ~> ~3.1§‘ﬁ71 0+(73V35‘1)At om: 03545
(5 PTS) (b) Determine the angle at which the frog Jumps. ' Va: :9} = 12.5.3: : (.133 /
At 0.6173 s \/ : 2,67‘; y}; imam above fart (a)
(“W““37 ....... ..
 q _, W; ’ _, 1675'?” Am I M6) =1 65$
QV‘ — 1mm {7;}  l (“,5 '75 V‘ ( L/ m1 (5 PTS) (c) What is the instantaneous acceleration at the instant that the ﬂy is caught? __£ 0i 3 1%?» dowvxwrd oY iay: 'ﬁr%ﬂv jﬂly: O“.
i _ ................. .me Page 6 6. A particle moves counterclockwise on the 50.0cmdiameter circle below. The tangential velocity v,
changes according to the graph below. OJoo' \
' . 0.0 / \
W?) 0.060 ,/
4) I
0.040 \ /
0.020 \
\ _ /
o O 7. 4 6 Jas) \5c (15 PTS) (a) Determine themagnitudes of the radial acceleration (in other words, the centripetal
acceleration) and the tangential acceleration at t = 3.00 5. Give both values.
RADIAL Acc ELeMﬂ 0N
’L o, o m 1
ac: Y3; :( 65,) ; 81.0””in r W v15; YA~MLQMEI 0.7.ng 0y Coozwg‘g? Mutter/ML ACCELEILATlon/ ,
1 At U 'ti ~ 2,; Li; (W 004521! clockwllf (5 PTS) (b) Draw three vectors on the circle above: the radial acceleration (in other words, the
centripetal acceleration) at t = 3.00 s, the tangential acceleration at t = 3.00 s, and the total acceleration
at t = 3.00 s. The magnitude and direction of the vectors should be sketched approximately to scale, so that the angle associated with the total acceleration is within 20 degrees of being exact. Wmmm_~w, ., W , Page 7 (20 PTS) 7. A cannon with a muzzle aimed 23.00 above the horizontal is used to start an
avalanche on a mountain slope. The target is 3000 m from the cannon horizontally and 800 m above the cannon. What is the magnitude of the cannon ball’s velocity as it leaves the muzzle? Ax : [\Vl cos Z3°>At :> At : AK : 3000M
/ V,~ ((33230 \I‘( (05130
Av = (v; wmc + am We) M Q New Wt +zn23° : 04124? $00M : [173m — V‘x ._ "3.10% we”
( <200 M — (173m ...
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 Spring '08
 Edwards
 Physics

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