2009f_1 - Physics 2101 Exam#1 Fall 2009 September 8 2009 N am e 5 i D I_m Section{Circte one 1{Chastain MWF 8:40 AM 4(Plummer TTh 9210 2(Chastain

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Unformatted text preview: Physics 2101, Exam #1, Fall 2009 September 8, 2009 N am e 5____________________________ i D# I___________________m_._______ Section: {Circte one) 1 {Chastain, MWF 8:40 AM) 4 (Plummer, TTh 9210) 2 (Chastain, MWF 10140 AM) 5 (Adams. TTh 12110) 3 (Rupnik, MWF 12:40 PM) 0 Please be sure to write (print) your name and circle your section above. - Ptease turn OFF your cell phone and MP3 playerl ° Feel free to detach, use, and keep the formula sheet. No other reference material is aliowed during the exam. 0 You may use either a scientific or a graphing calculator... I 0 GOOD LUCK! 1. (10pts) A car travels 20 kilometers at an average speed 0f 80 km/h and then stops for 15 minutes to get fuel. lt then continues on in the same direction for 20 kilometers at an average speed of 40 km/h in the same direction. The average speed of the car for this 40- km trip is about: (a) 40 km/h (b) 45 km/h (0) 48 krn/h (d) 53 km/h (e) 60 kin/h 2. (10pts) Consider the two vectors A = 5? + 4} and E = —3i + 103'. What is-the magnitude of C” = El — 3’? (a) 2.3 (b) 7 (c) 9 (d) 10 (6) 15.4 (Use this information for PrOblems 3 and 4) A model rocket is fired vertically from ground level, starting from rest. It ascends with a vertical acceleration of 5 m/s2 for 10.0 3. Its fuel is then exhausted, so it continues upward as a free-fall particle and then falls back down to the ground. 3. (Spts) What is the speed of the rocket when the fuel runs out? (a) 100 m/s (‘0) 50 m/s (c) 5 m/s ((1) -9.8 m/s (e) zero 4. (lOpts) What is the speed of the rocket 13 s after launch? (a) zero (b) 9.8 m/s (0) 20.6 m/s (d) 50 m/s (e) not enough information given 5. (lOpts) How far does the runner, whose velocity-time graph is shown below, travel between t = 85 and t 2 16s? (a) 64 m (b) 46 m at (c) 44 m E (d) 32 In (e) 10 m 3% 6. (Spts) A 10 kg block is allowed to slide down a long frictionless incline. The angle of the incline is 8 : 30 deg. If it starts from rest, what will its speed be after sliding 1.63 111 along the incline? (a) zero (b) 1.5 rn/s / (c) 3.3 m/s (d) 9.8 III/S (e) 4.0 m/s 7. (Spts) A football is thrown at angle of 50 deg into the air. Which statement below is true about the highest point of the football’s motion? (a) The football’s speed is zero. (b) The horizontal component of football’s velocity is zero. (0) The vertical component of football’s velocity is zero. ((1) The football’s acceleration is zero. (e) The magnitude of the football’s vertical and horizontal velocities are the same. (Use this information for Problems 8, 9, and 10) Block A in the figure below has mass mA = 8.0 kg and is sliding down the ramp. Block B has mass m3 = 4.0 kg. The coefficient of kinetic friction between block B and the horizontal plane is 0.50. The inclined plane is frictionless and at angle 30°. 8. (Spts) What is the frictional force on 1113? chnmfiessw rr'tassiess prziiey -—--~ H (a) 19.6 N (b) 39.2 N (c) 5.3 N ((1) zero (e) its equal to the tension in the string 9. (lOpts) What is the magnitude of the acceleration of the blocks? (a) 9.8 m/s2 (b) 5.2 m/s2 (0) 3.3 m/s2 (d) 1.6 m/s2 (e) it can’t be determined without further information 10. (Spts) What is the tension in the string connecting the blocks? (a) 39.2 N (b) 30.1 N (c) 26.1 N ‘ (d) 19.6 N (e) 13.] N - (Use this information for Problems 11 and 12) In the figure below, a stone is projected at a cliff of height h :‘60 m with an initial speed of 75.0 m/s directed 600° above the horizontal. 11. (lOpts) When does the stone strike the top of the cliff? (a) 12.3 s (b) 9.8 s (c) 4.3 s (d) 1.0 s (e) 0.5 s 12. (Spts) What is the speed of the stone just before it hits the cliff? (a) zero (b) 9.8 m/s (0) 55 m/s (d) 67 m/s (6) 38 m/s ' 13. (Spts) An astronaut on the space station whirls a 0.5 kg mass attached to a string in a circle of radius 1.6 m. If the speed of the mass is 8 m/s, what is the tension in the string? (a) 64 N (b) 32 N (c) 20 N (d) 4.9 N (e) none of the above 14. (Spts) Two forces are applied to a 2 kg object oating out in space. One force has a magnitude of 20 N and the other has magnitude of 30 N. Which of the following represent the minimum and maximum of the possible acceleration that the object can have, depending on the relative directions of the two forces? (consider magnitudes only) (a) 0.0 m/sz, 15 In/s2 (b) 0.0 111/32, 10 m/s2 (0) 2.1 rn/sz, 4.4 111/52 (d) 5.0 m/sz, 25.0 m/s2 (e) 0.0 m/sz, 8.3 m/s2 Formula Sheet for LSU Physics 2101 Exams, Fall ’09 Units: 1m : 39.4in = 3.28% lmi : 5280ft 1min = 608, 1day x 24h lrev = 360“ : 27rrad 1 K 9°F latm 2 1.013x105 Pa lcal : 4.131} T = (1°C) TC + 273.15K TF = (500)1"(3 + 32°F Constants: g = 9.8 111/32 REWh = 6.37 x 106mm MEmh = 5.98X1024kg G = 6.67X10‘11m3/(kgs2) RMM : 1.74 x 106m MMm m 7.36x1022kg Earth—Sun distance : 1.50x1011 1n MSW : 1.99x 1036 kg Earth—Moon distance : 3.82x108n1 k m 1.38 x 10‘23J/K R = 8.31 J/(mol-K) Avogadro’s # : 6.02 x1023 particles/11101 Properties of H20: 7 Density: pwater : 1000 kg/m3 Specific heat: cwater = 4187 J / (kg K) cice = 2220 J / (kg K) Heats of transformation: Lvaporization : 2.256 x 106 J /kg infusion : 3.33 x 105 J/kg ——b i x/bfi — 4 Quadratic formula: for £15132 + be: + c = 0, $1,; = —2——'—ac a Magnitude of a vector: |Ei| = 1m; + a: +n§ Dot Product: 63 - 3 = nmbm + aybg‘r + azbz = Iii] cos(q5) (qt is smaller angle between 51' and 1;) Cross Product: ('1‘ X 1;: ((1be «~— azbyfi + (c121)m — awbzfiwl— (1:11,,ijy — aybmflc, (a: X a = iii] sin(q5) Equations of Constant Acceleration: linear equation along 1: missing missing rotational equation vm=vow+amt m—zco 6—»60 wzwo—E—amt a: — $0 = vomt —£— gamtz 1:93 o: 6 — 60 = wot + E04152 1):; = 12333 + 2am(m -— 11:0) t t (112 = Lug + 201(6) — 60) 513—330: —:—(vom+vw)t am a 9—80: $(wo+w)t a: — $0 = 'umt — $11th vow mo 6 ~ 60 = wt — $05152 Vector Equations of Motion for Constant Acceleration: 1" 2 F0 + fiat + 56:12, 17' = 170 —£— at Projectile Motion: (with + direction pointing up from Earth) . 1 m — mo = (110 cos 60)t y — yo = (no sin Bo)t *— e2~gt2 ’03 = '00 cos 60 119 2 (no sin 90) — gt - 2 2‘ - . gm 1) sm 26 ) v: = (no Sln 9602 — 29(y w yo) y 2 (tan 90).:1: — m———2 R = LLm(—° 2(110 cos 60) g Newton’s Second Law: 2 E 2 m5: , _ . 711112 2 7r 1'- Un1form c1rcular motion: Fc = 2 mac T = r 1) Force of Friction: Static: f5 g fsmam = psFN, Kinetic: fk = nkFN Elastic (Spring) Force: Hooke’s Law F 2 —km (16 2 spring (force) constant) 1 Kinetic Energy (nonrelativistic): Translational K = Emvz Work: —o—; W = F- (constant force), W E f 951'. {E f 7f —o F($)da: (variable 1—D force), W = f - d1? (variable 3—D force) ‘7’ Work - Kinetic Energy Theorem: W = AK = K f — K: where W is the network ...
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This note was uploaded on 05/12/2011 for the course PHYS 2101 taught by Professor Grouptest during the Spring '07 term at LSU.

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2009f_1 - Physics 2101 Exam#1 Fall 2009 September 8 2009 N am e 5 i D I_m Section{Circte one 1{Chastain MWF 8:40 AM 4(Plummer TTh 9210 2(Chastain

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