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Exam2 solution

Exam2 solution - PHSX 114 Exam#2 code number 1 2 l(10...

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Unformatted text preview: PHSX 114 Exam #2, code number 1 2 l. (10 points) The vector E1 of magnitude 7.7 has an angle of 218.7° with respect to the positive x—axis (see figure). Find the x and y component of this vector. Ax=7~1* Coy-WT Z —' 6.0 c) %:7.0ff4.sjfi A721] ksmwgqo d) A=—6.4i—7.9j 218.?0 e) 21:43:29.2] 234.95 A7: Y < ‘1 A7. A‘ ~ *4 V 2. (10 points) Two forces 1?] = —(3.0N)f + (4.0N)j' and 17; = (6.0N)f + (7.0N)} are applied to a box. The magnitude of the total force on the box is: a 0. 1:: *4 ‘5. b) i4? v =F1+Fz : r Ffllgm‘i'F-ZXE‘EN‘F :BN‘ 6) 1.2.2N F7, :FW+ET1.4N+7N: l\.l\5. FnJFFQTFVL: Chill "2 H‘LILN' 3. (10 points) A car of mass 800 kg made a sudden stop in 5.00 seconds from its original velocity of 20.0 m/s in the positive X—direction. Assumin'gJa constant force was applied to stop the car, this force is: P ‘z’ 7 20‘ D M a) 1900 N, along the positive X—direction g j \J a: —~—_——é : arty/2. D b) 3200 N, along the negative X-djrection 5 5 ‘ c) 520 N, along the positive x—direction F : m CN :1 3 00 k5 x 4‘. M 1 d) 160 N, along the negative X—direction /3 A e) None of the above :1 a"; DO N 0408 __ . 4. (10 points) A force F acts on mass M for a time interval T, giving it a final speed v. If the same force acts for the same time on a different mass 0.5M, what would be the final speed of the smaller mass? _ fo—E-i , m :L: god a 4v ‘ 0.5M ' b) 2v Ul:alT_ c v d) M212 _~ —~ __ " 1‘ e) U412 D1—all H PHSX 114 Exam #2, code number 1 3 5. (10 points) Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the ratio between T1 and T3? aiTl/iii m g QnJ'mggmj o\ ,2 émcm. C)T1/T3:1/3 ._ d)T1/T3=l/6 T77 “ m 0" 6) T1 /T3 =1/2 T 1 3m 6. (10 points) A person of 100 kg mass is standing on a scale inside a moving elevator. What is his apparent weight showing on the scale if the elevator is moving with an upward acceleration of 4.9 mfsz? W N"VY1 :m&. a) 980N % 15 b) 490N \ TCK‘ W:m(é+&) _ 578N lei) 1960N I ROMS "(l'glékoU 2—3. I: [470 M» 7. (10 points) A child throws a ball horizontally with an initial speed of 15 m/s. The ball leaves her hand 1.2 m above the ground. What horizontal distance does the ball travel? a) 0.27m x yii‘lM: i512. b 1.8m 2\@ x.“ m “M \x t : .i’iii“: :oa‘iss, e) ism 3 firm gar all . $3 in ct FOL/moi L—lSMSKQQ—EZlLFm LS . 8. (10 points) An airplane is traveling horizontally at constant speed. A package is - dropped from the airplane. One second later a second package is dropped. Assuming air resistance effects can be neglected, which of the following is an accurate statement? a) The vertical distance between the two packages will remain constant as they fall. b) The horizontal distance between the two packages will increase as they fall. 0) The second package will hit the ground more than one second after the first hits. he vertical distance between the two packages will steadily increase as they fall. 6) The horizontal distance between the two packages will decrease as they fall. PHSX 114 Exam #2, code number 1 4 9. (10 points) A projectile is launched with an initial velocity of 60 mfs at an angle of 30° above the horizontal. What is its horizontal range over level ground? ‘2. a 180m M02 SCWZQ @O%)’gl‘w 600 0) 150m 6 2 M d)230m Chg} 1. e)400m /LS l1.3\8“.lm 10. (10 points) In the figure, the kinetic coefficient of the table is #13025. What is the acceleration a of the system (Box A + string + Box B)? ) 7.1 m/s2 H : T. m if m 4.0kng 1:) 3.3rr1/s2 Angjq AR 1:234/ 7 c) 1_4m/Sz E) : m5 3 - l :. m& 0\ g 2 ~~ 2/... W2) Afi'Eb ;‘ Millik"';::','""""“it” @b‘mt‘fll93=(mstma0\ W3 3 : “‘“L F. i .7 N: W. - j ‘2.qu a: W3; (6% w I?) W‘s-WW5 é Ffir’tN MK Extra credit arablem (10 Qaintsg: A 1400 kg car is traveling up a 15" slope at 28 m/s. The driver slams on the brakes and skids to a halt. If the coefficient of friction is pk=0.79, what distance does the cagravfi ‘0 7< before it stops? YA \J / iqu oilch‘i'Fo‘h . 5 7 a) 155m J >O~ 9%" b) 79m mflCosls * FM :0 J 0 0) 52m 8 "o d) 5111] 0 Five ijosl’D m3 F1”:MKFN \. m X —~0LireC/‘i‘j¢rv\_ m6 SinlSo +th5<losi5° 2 Wok. SO alé‘mlb—O ‘t‘ 0.79ACOEAS ‘fix‘ligm/sz :(O' O” ...
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Exam2 solution - PHSX 114 Exam#2 code number 1 2 l(10...

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