ME211_HW#10

Formal TL Gyab'G25 için 4 = 1-0,630b/a) Tmaks= Gab M.(lay 2-ly-y) + M,(4.2- lzy-y) .ly - ly Oy + dy tan20, - vQ 2tA cos20 + Taysin20, Jort 2 sin20 + Tycos28, omaks min- (의)'+다 Tmaks = +r A 3 m long steel angle shown in the figure has a section thickness of 12.7 mm and a cross sectional area of 4350 mm ^ 2. Find the resulting torsion angle in degrees. T= 500 Nm ve G = 77.2 GPa () O 5,34 O 3.89 O 4,87 O 7,86 O 8,72

I am unsure of how the equation got to the final result.  Can you elaborate on the steps between the last four arrows of this problum?

2. [-/1 Points] DETAILS KATZPSEF1 5.P.048.MI. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER To get in shape, you head to the local gym to exercise by lifting weights. Using a lat machine as shown in the figure, you notice that the wire connected to the bar is run around three different pulleys and is connected to a vertical stack of weights on the other end. The weights move up as you pull down on the bar. Suppose you set the machine so that the weight to be lifted is 124.5 lb. When you pull down on the bar, exerting a constant force, the acceleration experienced by the weights is 0.215 m/s2. what tension must exist in the wire connecting the bar and weights under these conditions? Assume the pulleys are massless and frictionless. N Pulley 1 Pulley 3 Pulley 2 Need Help? Read It Master It

WRITE SOLUTION HOW DID YOU GET FBD and FBE THANKS

Engineering Analysis. Plz help me help

!! X Quiz No.1 x Quiz No.1 docs.google.com/forms/d/e/1FAlpQLSe-TSkiDw1mbMifeQk9X100r-vpYjBy Q5- B) for the figure below If X1-20, X2-D30, X3-22.5, X4=10, X5-D0, X6%-D15, X7-20 and Y1= 80, Y2=65, Y3%3D60, Y4=15, Y5-D10, Y6-Y7%3D0 all in cm. Find: 0knee, 0Ankle and OMT-PH * 43 es OAOUO

Need hep with this Q.

Y(t) = t³e-t => Ys) =? A: 3 Y(s) (s + 1)+ B: 6 Y(s) (s + 1)3 C: 6 Y(s) (s + 1)+ D: 3 Y(s) (s + 1)2

Q4/ For the figure below:* a) gFy is: 0 0 0 0 0 0 C Op С O-860.12 O 0 0 0 0 0 0 C 0.6m 0.2 F2=1000 N -824.59 С -831.64 -900 -789.22 Q4/b) EFx is: * 0 0 0 0 0 0 0 - 840.88 Other: O-525.146 O-521.46 O -510.2 • -499.54 490.54 -515.32 0 Other: Q4/C) The resultant R is * 977.6 983.57 950.21 945.54 960.24 965.22 400 Other: 0.8m F1=1600N 0.4 4 F3=900 N 3 F4=1200N Q4/d) The resultant direction is 0 0 0 0 0 0 0 0 58 60 65 54 50 52 63 Other:

Solve the following differential equation: y'"-2y' +y= e*Inx Select one: a. YG.s=C;Xe¯*+Czx²e*+e* Jinx+ xe-* [Inx = C, sinx+ c>cosx+ sinx Inx+cosx /Inx b. YG.S c. None O d. YG.s=C,xe*+czx²e*-e* ] xInx+e* ]x?inx YG.s=Ce*+c2xe* - e* | xlnx+ xe* | Inx е.

Q4/ For the figure below:* a) gFy is: 0 0 0 0 0 0 C O С O-860.12 O 0 0 0 0 0 0 C 0.6m 0.2 F2=1000 N -824.59 С -831.64 -900 -789.22 Q4/b) EFx is: * 0 0 0 0 0 0 0 - 840.88 Other: O-525.146 O-521.46 O -510.2 C -499.54 490.54 -515.32 0 Other: Q4/C) The resultant R is * 977.6 983.57 950.21 945.54 960.24 965.22 400 Other: 0.8m F1=1600N +0.4 4 F3=900 N 3 Q4/d) The resultant direction is 0 0 0 0 0 0 0 0 F4=1200N C 58 60 65 54 50 52 63 Other:

Please provide a step by step process (in sentences) on how this was solve. Think of it as if you’re going to present this to the entire class

PLS USE THE ODD TABLE, MY LAST DIGIT NU IS 3

Q4/ For the figure below:* a) gFy is: 0 0 0 0 0 0 C 0.6m 0 0 0 0 0 0 C 0.2 O-860.12 F2=1000 N -824.59 0 0 0 0 0 0 0 -831.64 -900 -789.22 - 840.88 Q4/b) EFx is: * Other: O-525.146 -521.46 -510.2 490.54 -499.54 -515.32 0 Other: Q4/C) The resultant R is * 977.6 983.57 950.21 945.54 960.24 965.22 400 Other: 0.8m F1=1600N 0.4 4 F3=900 N 3 F4=1200N Q4/d) The resultant direction is 0 0 0 0 0 0 0 0 58 60 65 54 50 52 63 Other:

Q4/ For the figure below:* a) EFy is: 0 0 0 0 0 0 C 0.6m 0 0 0 0 0 0 C 0.2 O-860.12 F2=1000 N -824.59 -831.64 0 0 0 0 0 0 0 -900 -789.22 - 840.88 Q4/b) EFx is: * Other: O-525.146 O-521.46 -510.2 490.54 -499.54 -515.32 0 Other: Q4/C) The resultant R is * 977.6 983.57 950.21 945.54 960.24 965.22 400 Other: 0.8m F1=1600N +0.4 4 F3=900 N 3 F4=1200N Q4/d) The resultant direction is 0 0 0 0 0 0 0 0 58 60 65 54 50 52 63 Other:

help pls

A 令 Wed 17 Mar 12: moodle1.du.edu.om DU - Moodle English (en) - Mohammed Ahmed Al Sarj Al Marhoon tion 6 et ered K ed out of 000 ag cion M (t) fv Mä + f,* + kx + f(1) = 0 Mä + f,x+ kx = f (t) c. fyä+ Mx + kx = 0 d. fä+ Mx + kx = f(t) а. b. %3D The differential equation of the system is Select one: а. с O b. b O C. d d. a étv W acBook Air DII DD F7 F8 F9 F10 F11 F12 V 9. P ]{ 3 * CO

Tt) 02(1) 0,(1) N-m-s/rad I N-m-s/rad $ kg-m2 O3 kg-m2 3 N-m/rad 9 N-m/rad (a) Figure P2.1 lehn WieyA Sons Inc. Al ghts reserved The matrix form of the system is 5s² +9s+9 (s+9) T(s) а. (s+8) s?+s+1) 5s +9s+9 b. -(s+9) 3s +s+12)0,0 (0,() 3s +s+120,(s) 0,(s)) -(s +9) 5s +9s +9 с. (s+9) -(s+9) 5s +9s+9 d. -(s+9) -(s+9) 3s +s+12 ) 0,(s). Select one: a. b b. a С. С d. d

7:22 ull || 34% ENS 161: STATICS OF RIGID BODIES PROBLEM SET 1 Due: Oct. 20, 2021, 12nn Instructions: 1. Copy and answer the problems. Handwritten. Make sure you write legibly and orderly. 2. Take photographs of your solution. Please provide a readable photo. 3. Upload image files of your photographs as submission. Make sure you have successfully sent your files before leaving the page. 1 P= 70N at 0. and Q intersects Determine B if the resultant of these forces is 10ON. Also, determine the direction, 0, of the resultant. Forces P a = 40° Q= 50N 2 250 T2 B A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T, = 1000 īb, deter- mine by trigonometry (a) the required tension T¡ in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R. 1 of 6 A steel tank is to be nositioned in an excavation. Knowing that the ||

Class X Class k Ingrid x k Ingrid x C Auto: X © Shop X docs.google.com/document/d/1DUa0kRhTJwq|pEZJc3GcnvVhgTU09aBcA54WqeY7hMI/edit Supreme Court Cas... e New Tab ← → C D sd206.org bookmarks E A T Ingrid Lyons - Chapter fi... File Edit View Insert Format Tools Extens 100 I I I 100% I I Normal text ▼ I 100 I | Ingrid x Auto Orientation Radeke I I ~ Messe X Arial 2 2TO k Y Domi X I I 3 I I с 1 х 14 + I I I 4 G hown x I O I | Chapter review questions. 1. What noise level is too high and can hurt your ears? 2. What are the Safety Data Sheets? 3. If you get something in your eye what should you do? I TextNxb Home X 5 TURN IN I I | I I I 6 + Share I I I Ar 7 19 ☐ 31 > X :

TRUP TUE FLE FM AutoSave Hwk-6a.MechanialProperties-ElasticRegion • Last Modified: Mon at 9:05 AM - O Search (Alt+Q) Torialai Stanikzai On TS File Home Insert Draw Design Layout References Mailings Review View Help O Find - S Replace Times New Roma - 12 A A Aav Ao AaBbCcDc AaBbCcDc AaBbC AaBbCcC AaB AaBbCcC AaBbCcDa AaBbCcDa AaBbCcDa B COPY S Format Painter Paste BIUab x, x² A evA 1 Normal I No Spac.. Heading 1 Heading 2 Title Subtitle Subtle Em.. Emphasis Intense E. A Select v Dictate Edi clipboard Paragraph Styles Editing Voice Ed Font 5. Consider the difference between tension and compression. We defined mathematical expression for o, e, and v in tension. How are these expressions affected when applied to compression?

Q1 / Reduce the block diagram to the simple form R(s) S 13 S 0. S 1 S C(s)

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Formül; TL Csab³G' 525 için c = c2 =;(1– 0,630b/a) 60 MPa Tmaks = 45 MPa Cah? - $ = M-(!zy.z– ly.y) + M,(!,.z– Izy-y) Iz.ly – Izy² Ox + dy VQ Мy =1 It T= 2tA 100 MPa 100MPA Ox + đy đx - đy 2 -cos20 + Txysin20 , Jort 2 2Txy tan28, = Ox – dy Ox+ dy + 2 Ty'y' = Ox- dy sin20 +Trycos20, Omaks.min = + r 45 MPa 60 MPa Tmaks - Find the maximum principal stress for the stress case given in the figure. O 80,00 MPa O 138,75 MPa 129,24 MPa O 114,25 MPa O 49,24 MPa

please check my solution thank you

i want a displacement and stress equations from the global matrix that you can find it from the code and make a calculation %‏‏‎ ‎Define‏‏‎ ‎node‏‏‎ ‎coordinates‏‏‎ ‎(x,‏‏‎ ‎y)nodes ‏‎=‎ ‎[0,‏‏‎ ‎0;‏‏‎ ‎3,‏‏‎ ‎6;‏‏‎ ‎6,‏‏‎ ‎0;‏‏‎ ‎3,‏‏‎ ‎12;‏‏‎ ‎6,‏‏‎ ‎12;‏‏‎ ‎9,‏‏‎ ‎6;‏‏‎ ‎12,‏‏‎ ‎0;‏‏‎ ‎9,‏‏‎ ‎12;‏‏‎ ‎12,‏‏‎ ‎12]; %‏‏‎ ‎Define‏‏‎ ‎the‏‏‎ ‎connectivity‏‏‎ ‎of‏‏‎ ‎the‏‏‎ ‎elements‏‏‎ ‎(start_node,‏‏‎ ‎end_node)elements‏‎ ‎=‏‏‎ ‎[‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎1,‏‏‎ ‎2;‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎1,‏‏‎ ‎3;‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎2,‏‏‎ ‎4;‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎2,‏‏‎ ‎5;‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎3,‏‏‎ ‎5;‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎4,‏‏‎ ‎6;‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎5,‏‏‎ ‎6;‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎5,‏‏‎ ‎7;‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎6,‏‏‎ ‎8;‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎7,‏‏‎ ‎8;‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎7,‏‏‎ ‎9;‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎8,‏‏‎ ‎9]; %‏‏‎ ‎Define‏‏‎ ‎material‏‏‎ ‎properties‏‏‎ ‎and‏‏‎ ‎cross-sectional‏‏‎ ‎areaE‏‏‎ ‎=‏‏‎ ‎30e6;‏‏‎ ‎%‏‏‎ ‎Young's‏‏‎ ‎Modulus‏‏‎ ‎in‏‏‎ ‎psiA‏‏‎ ‎=‏‏‎ ‎1.2;‏‏‎ ‎%‏‏‎ ‎Cross-sectional‏‏‎ ‎area‏‏‎ ‎in‏‏‎ ‎in^2 %‏‏‎…

create matlab code and screen shot both files

(Q-4) Balance the meter stick with two weights and a rock. 0 cm 100 cm Jrock 100g position (cm) weight (gwt) lever arm (cm) torque (gwt cm) Rock 10.60 XXXXXX XXXXXXXX Weight 1 구군, 60 150 cw Weight 2 93.5 100 (Q-5) Assume that the torques balance and use this fact to calculate the weight of the rock. Z Tecw Złcw = Wrack

I just need help simplifying these equations (1 and 2). Qo,Kic, and CL are all their own variables (not 2 multiplied ie. Q*o, k*i*c and not C*L)

Draw the corresponding scheme and find the equation on the right. Method used and Four assumptions hA Tc.v.(t) = Taire + [Tc.v.(tt) – Tairele locp)"

Step 3