shaded area. Ans. X 244 mm, Y = 117.7 mm 200 mm - 200 mm 150 mm 125 mm 60 mm 125 mm

Q: Locate the centroid y of the shaded ared. L=23.6mm y 2L L 2L 2L

A:

Q: Question 240K-ft where E = 1800 kst H=13,432 int 24 Ft Cal culate andl draw the de formed Shafe.…

A: Given  E = 1800 ksi  I= 18432 in 4

Q: LINE LATITUDE (m) DEPARTURE (m) АВ 100 50 ВС 200 200 CA -300 -250

A: Calculating the DMD(Double meridian distance): Double meridian distance of any line=DMD of preceding…

Q: 1. Refer to the setup and data below. Determine the length of side AB. Stadia interval factor: 100…

A: *Multiple questions*   As per our guidelines we are supposed to answer?️ only one question. Kindly…

Q: 152*102*11.1. Use

A: we have to find the net area

Q: 3/17 4 m 8, 30 Y--- x 300x9.81 N A

A:

Q: Determine the y-coordinate of the centroid of the shaded area. y 0.30h 58° V 58° -x

A: From the geometry,tan32o = ABh ⇒ AB = 0.62hSo, AC = 0.62h + 0.62h = 1.24htan32o = EF0.3h ⇒ EF =…

Q: Find the centroid along x- axis of the shaded area. x - axis mm. O 135.4 y O 183.7 O 169.8 O 164.6…

A:

Q: Determine the y-coordinate of the centroid of the shaded area. y h 0.27h / 55° V 55° - x

A: To find out the y co-ordinate of centroid of shaded section.

Q: X= 482  Y=82 C=2 SHOW COMPLETE

A: X=482 Y=82 C=2

Q: Compute the Interior A. AB= N 60° 21’E BC= S 48° 31’E CD= S 16° 20’W DA= N 63° 41’ W

A:

Q: Given the multiview projection of the solid, draw its isometric view. HRP FRP FRP PRP ㅏ

A:

Q: Given: Lines Bearing S 73°21'E S 40°10'E S 26°42'W N 14°20’W N 12°20'E Distances А-В 247.20 В-С…

A:

Q: 0 mm 20 mm x= ky? 30 mm etermine: enter of Area in x direction, overline x 1U-Sclect) mm enter of…

A: To find centroid

Q: 20 kN y 30 -x- 45° 8 kN 4 kN Problem 2/80

A:

Q: 2m y 4m 5m 2.5m X 10m

A: To Determine The Centroid of the shaded region (x-bar and y-bar) Divide the whole region into 4…

Q: Determine the x- and y-coordinates of the centroid of the shaded ar ! 3", 3" 4" 1.25" 10" 1.25" 3" B…

A:

Q: Problem 9-12: y-axis 90 mm 100+ V (mm) 60 mm x-axis

A: Given Data:X=137, Y=37, A=6, B=8, C=1 The solution to the above given questions are as under-

Q: Determine the y-coordinate of the centroid of the shaded area. 0.33h / 61 61° Answer:J =

A:

Q: Locate the centroid x̄ of the shaded area.

A:

Q: Determine the y-coordinate of the centroid of the shaded area. y h 0.35h Ł- 62 V 62° Answer: y =

A:

Q: Azimuth in DMS from Point A to Point B: 45-14-37 Distance from Point A to Point B: 927.56' Azimuth…

A: Fig.

Q: how far is the center of the circle x^2+y^2-6x-8y-11=0 from the y axis asap pls

A: Express the equation in the general form. Then compare to find its center

Q: find the elevation of reduced level for all point For sight Reduced level Back sight point Inter…

A: Given that:

Q: F 31 ° 58 ° X(+) B=(3.3,0,0) m y(+) Z(+)

A: Given:- A coordinate system is given The coordinates of point B = (3.3,0,0) m  Point A lies on the…

Q: y 74 mm 36 mm - 36 -+ 36 mm mm Answer: y = i mm

A:

Q: 37 a = 6 b = 8 c

A: TAKE MOMENT ABOUT G∑ MG=0Ax(4+1)-137(3+8)-37(3+8+3+8)-37(3+8+3+8+2+8)=0Ax=701kN

Q: Figure 1 0.30m/s 0.05m2/s 1500m, 300mm 1500m, 300mm B 1500m, 300mm D A E 1500m, 300mm F 1500m. 300mm…

A:

Q: Compute the Interior B. AB= N 60° 21'E BC= S 48° 31'E CD= S 16° 20'W DA= N 63° 41’ W

A:

Q: Using Centroid by Integration, determine the x- and y-coordinates of the centroid of the shaded…

A:

Q: Find the centroid along x - axis of the shaded area. x - axis = _in. O 2.35 O 2.49 7.5" O 3.66 4.5"…

A: solution consider this figure into area of semi circle and area of triangle…

Q: Given the multiview projection of the solid, draw the isometric view. HRP FRP

A: Introduction- A two-dimensional depiction of a three-dimensional object or space is called an…

Q: 216 Home Works Find the centrod for the Shaded area. [Q2] :- 120 150 30 90 130 4

A: Given a areas ,we have to find centroid for the given area

Q: Find distance x at which head is 33.2 m 1285.24 .A O 1028.19 .B O 0.00 .C O 822.55 .D O

A: At X=0 h1= 45.7 m At X= 1690.9 m h2= 21.6 m Change in elevation = 45.7-21.6 =24.1 m Change in…

Q: A rectangular section with b = 250 mm and d = 350 mm, if f = 25 MPa and fy = 420 MPa, the depth of…

A: Given data Width of the beam, b= 250 mm  Depth of beam, d =350 mm Tensile strain = 0.006 fy = 420…

Q: the centrold of the following shaded areas 1. Use R = 60 mm. 60 mm R 1.5R 75 mm 45 mm 54 mm

A: To find centroid

Q: Pa, I = 1660 x 10 41KN/m, L = 14m three decimal pla

A:

Q: Compute the Interior C. AB= N 60° 21’E BC= S 48° 31’E CD= S 16° 20’W DA= N 63° 41’ W

A:

Q: TT 3.2 in. |B D 5.8 in.

A: Given:- P=7500lb Q=3000lb a=15ft b=8f c=10ft Iz=110in4 To find:- Magnitude of maximum tensile…

Q: Determine the x- and y-coordinates of the centroid of the shaded area shown. | 3". 3" 1.25" 5" 1.25"…

A:

Q: Given the multiview projection of the solid, draw its isometric view. HRP FRP FRP PRP

A:

Q: SITUATION 8 Find the head light distance for a speed of 65 kph. Given f = 0.36, t = 2.5 seconds %3D

A: Given data : speed (V)=65 kphCoefficient of lateral friction (f)=0.36Perception time (t)=2.5 second

Q: Determine the y-coordinate of the centroid of the shaded area. y h \9.5hl 61° 61°

A: area of the triangleA1=212×base×heightA1=212×hcot61×hA1=0.554h2height of the centroid of the…

Q: Locate the centroid of the following shaded areas. Use R = 60 mm. 60 mm 1.5R 75 mm

A:

Q: Determine the y-coordinate of the centroid of the shaded area. y 0.35h 56 V 56°

A: Tan 340 = B/2h B/2 = Tan 340 ×h Base of triangle ABE = (B/2) + (B/2)…

Q: Problem No. 3 Given: Bearing S 73°21'E S 40°10'E S 26°42'W Lines Distances А-В 247.20 В-С 154.30 C-D…

A:

Q: Determine the centroid of the shaded area. 200 500 10 200 All unit in mm 150 400

A: As given in the question a compound figure with a rectangle, a semicircle and a circle. The centroid…

Q: y 100 mm 140 mm 50 mm 140 mm -200 mm

A: To find out the centroid of the shaded section.

Length and Length related Variables

Length can de defined as the measurement or extent of physical quantities or objects from one terminal to another terminal. Stone age humans used finger length, step length, and so on to measure the size or displacement of an object. Nowadays, depending on how precise the measurement requires to be and the object’s size is measured, we use measuring devices for measurements like rulers and steel tape. Most people have used a ruler or tape measure to measure a particular distance or the size of an object. For more precise measurements of tiny objects, the development of measurement tools such as the Vernier caliper, which permits us to measure dimensions within 1/1000 of an inch. In the last few decades, electronic distance measuring instruments (EDMI) have been implemented that permit us to measure distances from a few feet to many miles with reasonable accuracy.