The second moment of area about z-axis of the cross section, Izz ( Units: mm4) can be calculated asSelect one:O 1. 2.081 x 107○ 2. 2.281x 107○ 3. 2.201 x 107○ 4. 1.348x 107○ 5. 1.466 x107○ 6. 1.966x107O 7. 1.833 x 107O 8. 1.109x 107

Answered: The second moment of area about z-axis…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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The second moments of area about z-axis, /z, and the second moments of area about y-axis, ly, can be calculated as Select one: O 1. I = Iz ○ 2. Ly ○ 3. ○ 4. ○ 5. = = Iz = *D' 64 I₁ = D, Iz Ly Ly = 32 *D' = = 3 Iz = *D' 32 = *D' O 6. Iy=D, Ly = D², Iz = 32 O 7. Ly = Iz D = 64 32
Calculate the Maximum X distance on the drawing below: 1.000 .510 O 3.000 2.009 O 1.996 2.000 с 2.000 Ø.995-1.000 6.00 0.008 MABC X 3.000 Y 01.005-1.030 +0.005MABC B A XX = .01 ANGLES= ± 1°
The following diagram corresponds to questions 1 to 3. A beam rests on two sharp edges as shown on the image. The beam has a length of 7,50 m and a mass of 4,25 kg. Object 1 has a mass of 1,50 kg; object 2 has a mass of 2,35 kg. Point P is 0,50 m form the center of the beam. CG 1. How much is the torque done by object 1 around point P? A. 55,2 N*m B. 7,36 N*m C. 62,5 N*m D. Object 1 doesn't exert a torque. 2. How much is the torque done by the force of gravity of the beam around point P? A. 20,8 N*m N-m ך177 .B C. 156 N*m D. 313 N*m 3. If you needed to cancel the nomal forces of the two objects, where you should place object 2? The axis of rotation is point P. A. 3,30 m from point B. 9,18 m from point P C. 5,69 m from point P D. 3,62 m from point P
For the area shown below defined by the two curves, to find Iy by integration you use a dA element parallel to the x axis. The expression in the integral is the following: y = x 100 mm 1 100 - X. 100 mm O ly²(10y1.5 – y)]dy O ^(10y².5 – y0.5)|dy O [x?(10y0.5 – y)]dx O [x°(10y0.5 – x)]dy O ly^(10y0.5 – y)]dy
need help engineering of materials
QUESTION 4 Find the resultant vectors. This referes to Question 4 on the PDF v Horizontal 3ON and Vertical 60N A. 67.1N @ 116.6 below the horizon v Horizontal 3ON and Velucal -60N B. 67.1N @ 63.4 above horizon v Horizontal -30N and Vertical 60N C. 67.1N @ 243.4 from the horizon v Horizontal -30N and Vertical -60N D. 67.1N @ -63.4 below the horizon
Three forces intersect at point (1, 1, 1). Their magnitude and respective second point of incidence are as follows: 450 lbs (4, -3, 7), 600lbs (-1, 5, -4), 300lbs (-3, 6, -2) 1. Compute the direction cosine on the x-axis. a. 0.394 b. 0.124 c. 0.762 d. 0. 513 2. Compute the direction cosine on the y-axis. a. 0.394 b. 0.124 c. 0.762 d. 0. 513 3. Compute the direction cosine on the x-axis. a. 0.394 b. 0.124 c. 0.762 d. 0. 513
Find the first moment of area, why don’t you include the middle portion in the calculation?
Let the region R be the area enclosed by the function f (x) = x³ – 1 and g(x) = 2x – 1.Find the volume of the solid generated when the region R is revolved about the line y = -3. You may use a calculator and round to the nearest thousandth. 2 1 -0.5 0.5 1 1.5 -1.5 -1 -2 -4 -5 -6 -7 -8 6- 2.
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1 Required information Consider given force vectors. Take F= 15 lb. y Q = 60 lb 30° 20 60° 60° 45° P = 100 lb Write an expression for each force using Cartesian representation, and evaluate the resultant force vector. = (- 4.58 i + 26.83 12.6 k ) lb 39.8 i + 30 j 33.4 k) lb 70.7 50 k) lb 50 85.22 i 13.87 29.2 k) Ib %3D
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The second moment of area about z-axis of the cross section, Izz ( Units: mm4) can be calculated as Select one: O 1. 2.081 x 107 ○ 2. 2.281x 107 ○ 3. 2.201 x 107 ○ 4. 1.348x 107 ○ 5. 1.466 x107 ○ 6. 1.966x107 O 7. 1.833 x 107 O 8. 1.109x 107