# Yellow Test Answers - Answers(YELLOW test}I m da 1hr 1min m...

• Homework Help
• PresidentHackerCaribou10582
• 3

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Answers (YELLOW test}I m da 1hr 1min — m 1105 points} a = 4 me’is dag) = —— = 5x10 5— = 4.7x10‘6 "g" s dag 24 hr so min so 3 s?- V 2] (10 pts} distance traveled = (ave speedﬂtime) or AX =‘Lgu—D-t or 0.05 Wt 3] {15 pts) a)LetT"1 represent one of the beams and-r'z the other ﬁ' bg the Pythagorean theorem: [HI = [le = 5!? m “—_— 7.2 71 T": o'r‘z = (53:15» (5)(-5)+(5}(5) = 25 m2 also: 7’1 or} = [F1 I ﬁzlcosqa = [5f312cosqj = 1'5 costp rn2 -. coscp = 25275 = 13 or tp = cos"(1f3}= 70.50 b) (10 pts} Just make this a vector problem, as I did in class with mg "hike around a circular path." Let ﬁbe the edge of the brace. From the diagram above: 7'2 =_r°1 + E or ﬁ' =‘r’2 —'r'1 = (5,-5,5) - (5,5,5) = (0,—10,o) or R = Iﬁ'l = 10 m 4] (10 pts} This is essentiallg the extra credit problem (which I also did in class). At time =T the bug is at location :4 = U’zaT2 and moving at speed v = aT Now the bug decelerates and its speed drops to 0: V v2 — v02 = 2a(x - x0) or 0 - (aTZl2 = 2(-a}(x - 1X2aT2} solve to get x = aT2 5] (20 pts} starting conditions: for No—No: x0 = 0 v0 = 10 ftfs a= 0 for Liberal: x0 = 30 ft 'v'g = 0 a = 2 ftx’s2 equation of motion f or No—No: x = 10t equation of motion for Liberal: >4 = 30 + t2 when they meet: XN = xL or 10t = 30 + t2 solve to get: t = [10 1 J-20 11'2 = imaginarg => theg never meet, the Liberal is spared! 6] {15 pts each} a)g = gg+vogt+1f2agt2 or 0 x= xo+voxt+1f2axt2 or' x 500 + 0 —1:"2gt2 solve to get t = 10 sec 0 +(15'130'1t1I + 0 = 1500 m b) vx = constant = 150 ms’s and v9 = v09 - gt = 0 -(9.EI}(10} = -98 mJ’s therefore? 150? -98’j‘ mfs when bomb hits the ground Name- A SHOW YOUR WORK IF YOU WANT CREDIT Here are some conversion factors: 1 inch = 2.540 cm; 12 inch = 1 foot; 1 mile = 5280 ft,- 100 cm = 1 meter; 1 slug = 14.59 kilograms; 211' radians = 360° = 1 rev; "9" = 9.8 mills2 = 32 fte's2 V1] The Japanese satellite Hagabusa is now in solar orbit and will rendezvous with an asteroid named Itokawa. The satellite‘s ion engines can accelerate it at a = 4 meters per second per dag. How does this compare to the acceleration "9" that you are used to? ' .. s l 9 “£2723, > __ A? 4’ /D m l—n/s '1 @— /y%‘40? ‘5 , 4 WWWMMWW 5 2/ £00 ' a @ 5 C73 ._ 21(éé3 MM” M a \ >‘\UJ‘ c4» 5 3 r I D’ 1/ < \ ¥ ‘ ‘ W'”"‘"“ W““J”Mm§4w:mwrmnN\»‘.‘.AAIWMINN\WWMﬂw>mmmmmwhuwm.mmum,*q in” UN ‘Ye.,,,_,.N,w.,v.u.w,wu. ., mutiny”. w— vvvww'v.w¢m:mw.»~~me-—m~:mwﬂs;.a\ a r as a w W A m w we mo a w l 0 ‘ \ 5W 1 W 4'94“ 65"“ 47C 9 f“ WW"— n M "why v, “mam”, . .ww.._m_‘___m_ 2] A bullet is fired completelg through a board that is 5 cm thick. The bullet enters at speed 200 mfs and emerges at 100 mfs. How long is the bullet in the board? D . . . . . . . . . . . . .. V 0. O§ I 0 + 200(S';>+ gl‘a(g. :: I00 0 05 61"” 2 is» Q __ A DVD + J” 0i , ﬁDDA ,g 0 " ﬂ + gﬂz 7» < 800* ’3 0.03 A Maw villa 1' 80’ a 2‘ ZUUZSm/sz /’Q/ ’ JDGAAFW ’3 rallying at? 5] Republican Captain hip—No spots a libeggding 30 feet away who is trying to register voters. Captain No-No wants to stop this and runs toward the liberal at a constant 10 ftfsec. The liberal sees the danger and flees on his solar—powered scooter which has an Vacceleration of only 2 ftISZ. What happens? m Now» lye/2 2!” 5‘ A”; I 415 30474- X2 3 X02 «- V2+ (@ (“‘4‘ “AA” : o + ID4- Xk: /O‘f 0‘ ‘ Email. We 9"‘3 b‘ ° etc/«whorl; o Mo4fl‘ﬂ ‘4'" \_/ 56L Q @14sz POSIJ-IOA a“?! /— m) H- wou’d 4"?“ 6’7“ ...
View Full Document

• Spring '06
• staff

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern