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3 years ago

What does WCS stand for? A. Western CAD System B. Worldwide Coordinate Sectors C. World Coordinate System D. Wrong CAD Settings

Engineering

1 answer:

Ray Of Light [21]3 years ago

7 0

Answer:

The correct answer is C. World Coordinate System

Explanation:

The World Coordinate System has to do with that coordinate system which is fixed in the activities of the CADing. There is a default system in which we can refer to them as soon as we want to manipulate the objects and add new elements.

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A thin-walled cylinder of average radius 50 mm, and wall thickness 1.0 mm is of length 500 mm, and is built into a wall at one e

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True or False; The Neutrons in an atom have a neutral charge.

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Answer:

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3 years ago

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Harvey has an inclination towards astronomy and wants to construct telescope mirrors. Which metal should he use for reflective c

lys-0071 [83]

Answer:

The metal Harvey should use for reflective coatings in telescope mirror is;

C. Aluminum

Explanation:

In a telescope, the mirror of the telescope is usually made of a temperature-resistant and strong glass, while the reflective coating of the mirrors is made usually from aluminum. A protective coating often composed of silicon dioxide is placed n the top of the reflective coating

Aluminum is used in a telescope, rather than the silver, which is the most reflective material, because silver requires special conditions to be able to work, and silver tarnishes when exposed to air containing sulfur, and silver is susceptible to corrode.

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3 years ago

A ramp from an expressway with a design speed of 30 mi/h connects with a local road, forming a T intersection. An additional lan

hram777 [196]

Answer:

the width of the turning roadway = 15 ft

Explanation:

Given that:

A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road

Using 0.08 for superelevation(e)

The minimum radius of the curve on the road can be determined by using the expression:

$R = \dfrac{u^2}{15(e+f_s)}$

where;

R= radius

$f_s$ = coefficient of friction

From the tables of coefficient of friction for a design speed at 30 mi/h ;

$f_s$ = 0.20

So;

$R = \dfrac{30^2}{15(0.08+0.20)}$

$R = \dfrac{900}{15(0.28)}$

$R = \dfrac{900}{4.2}$

R = 214.29 ft

R ≅ 215 ft

However; given that :

The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.

From the tables of "Design widths of pavement for turning roads"

For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation

Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.

As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;

If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft

Hence; the width of the turning roadway = 15 ft

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3 years ago