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Unformatted text preview: PHY 2425: Engineeringr Physics 1 Exam MB: 24 September 2501 _' —.___—_....__— —,_—,_._— _—...._—. _———_— I. A pirate ship is rnoored 520 m from a Fort defending a harbor entranco. The harbor defense consists of a cannon, located at sea level, with a muzzle velocity, v, = 94 l'l'lfs. a r I I 14:11
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in ﬁg”) / 1 WI H (5‘5
R=520m 19L (“Wt”) a IT
1231 : "314,:
.a} At what angle[e}, Finland 301, must the cannon be elevated above the
horizontal, in order to hit the pirate ship? 1.
:ﬂclnlﬁi H, Sir1+3: ﬂ 4 smith '3 Un1 fit—l “13331
,Gqﬁ mygu. 2.6 = no (“ W ') : 35.2 °.
Qinlﬁl : —._._____.. : 9 Lb) What are the Time of Fit'ghis, t] and t; , for the two {2) elevation angles calculated above? ‘
T we we. Ma‘m’g fairing) F
I —_ ___.__—_. :: ———————— — C} QKWIQ‘
. ' 2'.
3 ‘ r} ' 9.3mch Le) What are the Maximum Heights, H. and H1 , for the two {2] elevation angles
calculated above? '3.
1 1 e 'L r r
(U5 gin ﬁll) ( qLiFl‘ﬁ/S 3.5 gm 11.9 )l : x Mtg) ‘ H: :_————————'_'_‘“~ a D23 Maw/:1) we 2.) A pitcher tesses a hall ef mass, 111 = [1.21110 kg, Vertically straight up with an
initial velocity given by in; = 35 rats. la} How much time does the balI take tn reach its highest point? D UMﬂﬂf—fa t=£
reenttg M'— U —‘r We — ‘3 , g
i = 55 m” s 2,. rs s
ct Er "Us"
in: ZQWI‘L
250;”
lb) Haw high dues the ball rise above its reiease point? “If _ iii'HLC'il
Li ﬁleden 2,; [£11320 J HID Li : 0 Jr \iu'i — q t: gem/E (3,6g)_ L (agtﬁélytn) ini .51
L13 : rlgm— 633th : 62.5
1:) Convert, in} = 35 mfs tn rnii'hr.
aw i. “""i , ﬂ .3
E “twain” I ha 1d) If the ball were twice as massive, m = 9.41m kg, how much time does the ball
take to reach its highest pnint assuming same initial veleeity, w; = 35 rm‘sf? ‘ 1.
O— I   dig Til9 “Erma int"1e ULJl'hE'] Vitamins? 3.} Spotting a police car, you brake a Porsche from T5 kmﬂtr to 35 kmr’hr ever a
displacement of [25 m. Hint. convert tn SI Units [rate] ﬁrst!
la) What is the acceleration in SI units.1 a, which is aasmnetl to be eenstant? Pay attentientn Sign! _ +3—7me V : 30.?”93
«I: 11.31 +2di‘Krd) v a ctTt‘nre — 10me5 4— RELOL‘LBW) _ WWWBEDEW‘ili _ m 1 f  W /*=~ i w t " ngl 3.13} What is the elapsed time (t) uaed tc cover the dispiacepi’ent of I25 m? /
Jr : an an _ aTmi’s' meref3” 3c "Iu =129m H: “In _., a O. LL'W'IHL ri_‘gﬂf‘ .M ,_
N (in: 3.c] What graph depicts the above acceleration precess‘? Circle graph. In;er it; [{5} 4.] ‘I‘here are twn {2} veetnrs a and is given by a=3.0i+5.ﬂj /
b=lﬂ.E}i4.ﬂj ,, ..J 4.3.] What is the resultant veeter r = a + h in enntpunent farm, rx and r5, ‘? M’ r: ring = 030+ it: ﬁt + (5.0 + mum : ﬁll/4 13/? ra4'rtt/
4J3) What is the magnitude. r, ufthe resultant veeter r = a + h ‘i' ' '4. 'It 1  '0 : 1"]
Ft : \lﬁtﬁ, 111*“ “ E [:1 4.1:] What is the direction. 9* with respect to the xaxis of the resultant VEEIDF r '3 / t D
a (is) : lg...,q~(;~) : swap l
_ . {1 4d) What is the resultant 1treetnr s = a r h in enmpunent farm, a}, and s}. ? g _: (hr54:. ﬁlgyl + [:5 — (it?) ._ (41%E)A '31 4' 33/ 11 4e] What is the magnitude, s, of the resultant veeter s = a + b ‘2’ / A 1 — ll : '5' 9
IS! = VE'B) + 63‘ : tiles 4st  50
4.1“) What is the direction, EL with respect tn the intaxis (if the resultant veeter s 'i' . t ‘3 I
f] : tm‘1( iii) : inn" 3) : tam [— Uréqz)
St El 4 ﬂ: _31+7 5.) A car cf mass 25.!) kg and tl'aVcIing at [SJ] rat's attempts to make a left turn
with radius 60.0 rn an a level read at an intersecticn.
5.91) What centripetal acceleraticn, at ., is required ta keep Ihe car in its circular path? 5J3) What centripetal aeceleratien, a.“ is required in part 5a) ta keep the car in its
circular path ifthe car‘s speed is suddenly.I doubled (Le. 30.0 mfs) ‘? ZEICELMjL: JEN/Se / [at ﬁn m 1,3 ,
L 5.c] What centripetal acceleration, as. is required in part 5a] to keep the car in its
circular path if the car’s mass is suddenly doubled [i.e. EDI] kg) ‘? {IL a: qlWahml ’ ...
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 Spring '08
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