Solve ot A 30 m high vertical column of a fluid of density 1878 kg/m' exists in a place where g = 9.65 m/s. What is the pressure at the base of the column. An KA4 LD. Q1.6

Q: 5. 4

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A: We are given: 1/2×10 ft   The given expression is equivalent to computing half the length of 10 f.

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A:

Q: Find the centroid of the following cross-sections and planes.

A: Centrodal distanceX = 5.333 inY = 5.667 inExplanation:Step 1:Step 2:

Q: 1. GIVEN: REQOIREO: yo l0". 7,9 to locete the centreid t 6"

A: To find centroid

Q: Determine the ỹ coordinate. y b. = ky? a

A:

Q: Find the centroid for the shaded area

A: Given Diagram with dimensions    To find  Centroid of the shaded area

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Q: A C L- (2) B "1-12- -L/2→ C x

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Q: 1⁄6 yard +3⁄8 yard

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Q: midpoint approximation

A: Midpoint Approximation: The midpoint approximation or the midpoint rule is used for calculating the…

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A: Labelled diagram of the different section -

Q: 6"

A:

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A:

Q: 31⁄5 / 4⁄7

A: 31/5=6.2 4/7=0.571

Q: Q1: determaine the coordinates of the centeroid of the shaded area 20 40 50 20 40 30 140

A: The polar coordinate system is a method to calculate points in the system location by proper…

Q: نقطتان )2( Shows polygon with the followng :Find the area (x,y) 2 1.8 1.6- 1.4- 1.2 0.8 0.6- 0.4…

A: answer is option a.)   Matlab command window picture is attached below

Q: Convert 36° 19' to decimal degree form.

A: Given data: θ=36∘19'

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Q: Determine centroidal coordinates of the volume.

A: x¯=∑χiVi∑Vi   =610×660×6102 + 12×610×660×(610+13×610)610×660 + 12×610×660   = 474.45 mm y¯=8132    =…

Q: Y

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Q: 4 cm 0 2 י 5 cm 4 567 Value: cm 10 5 150 Value: mm 0 140

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Q: Prove that Ox - Oy sin 20 + Ty cos 20 and 『;-の cos 20 - Ty sin 20

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A: Given data: Area of each member is 800 mm2.

Types of Composite

Composite materials are formed by merging two components that have different chemical and physical properties. The merged output is termed as a composite, whose properties are completely different from the individual components. Composite materials also have better strength than the individual components.Composite materials usually have two main parts, reinforcement material, and matrix material. The matrix material acts as a binder and surrounds the fiber of another material that bears a much higher strength. This material is known as the reinforcement material. The matrix materials are usually high-strength additives resins, while the fibers of the reinforced material can be random, oriented, and in a mat format.Composite materials are also known as fiber-reinforced composite polymers because they are composed of the polymer matrix which is reinforced to laboratory-created fibers or naturally occurring fibers, such as glass fibers, carbon fibers, or aramid fibers. Under the application of external loads, the matrix material absorbs and transfers the loads to the fibers, these fibers in return provide good compressive strength, tensile strength, and stiffness to the fiber-reinforced composite materials.Depending upon the matrix and the reinforcement material, different properties can be induced to the fiber-reinforced composite materials, including good corrosion resistance properties. The most common matrix material is the resin while the glass fibers act as the reinforcement material. Strong fibers provide strength and stiffness to the composite materials while, a soft and flexible resin provides shape to the composite materials.