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3 years ago

What is the electrical cable that may only be installed in a damp or wet location

Engineering

1 answer:

uranmaximum [27]3 years ago

6 0

Answer:

Standard Armored cable(AC) or Metal-clad cable(MC)

Explanation:

Standard Armored Cable (AC) or Metal-Clad Cable (MC) may be used based on the definition of words "wet or dry locations" in areas that may be temporarily exposed to the elements such as rain during construction.

Dry locations should not be confused with damp locations which are protected from these elements and water saturation or moisture. Some damp locations may include open porches, barns, under canopies, etc.

Wet locations occur in places where direct burial in the ground, in concrete, etc may occur. These are places where water or other liquid saturation is very possible, or in places that are always opened to these elements.

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A particle is moving along a straight line with an initial velocity of 6 m/s when it is subjected to a deceleration of a- (-1.5v

Juli2301 [7.4K]

Answer:

s= 6.53 m

t=3.27 s

Explanation:

velocity = 6 m/s

deceleration = -1.5 $v^\frac{1}{2}$

$a=-1.5v^\frac{1}{2}\\v\frac{\mathrm{d} v}{\mathrm{d} s}=-1.5v^\frac{1}{2}\\-1.5ds=v^\frac{1}{2}dv\\\int\ {-1.5} \, ds= \int\ v^\frac{1}{2}dv\\-1.5s=\frac{2}{3}\times v^{\frac{3}{2} }$

now inserting value of v=6s we get distance(s)

s= 6.53 m ( distance cannot be negative)

now for time calculation we know that

$a=\frac{\mathrm{d}v }{\mathrm{d} t}$

$-1.5v^\frac{1}{2} =\frac{\mathrm{d}v }{\mathrm{d} t}\\-1.5dt=\frac{dv}{v^\frac{1}{2}} \\\int -1.5 dt=\int v^{-\frac{1}{2}}dt \\1.5t=2v^\frac{1}{2}\\t=\frac{4}{3}v^\frac{1}{2}$

putting value of v=6s

t=3.27 s (time cannot be negative)

3 0

3 years ago

A force is specified by the vector F= 160i + 80j + 120k N. Calculate the angles made by F with the positive x-, y-, and z-axis.

Rashid [163]

Answer:

1) Angle with x-axis = 42.03 degrees

2) Angle with y-axis =68.2 degrees

3) Angle with z-axis = 56.14 degrees

Explanation:

given any vector $\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}$

and any x axis the angle between them is given by

$\theta_x =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{i}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{x\cdot i}{\sqrt{x^2+y^2+z^2}} )$

Applying values we get

$\theta_x=cos^{-1}(\frac{160}{\sqrt{160^2+80^2+120^2}} )=42.03^{o}$

Angle between the vector and y axis is given by

$\theta_y =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{j}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_y=cos^{-1}(\frac{y\cdot j}{\sqrt{x^2+y^2+z^2}} )$

Applying values we get

$\theta_x=cos^{-1}(\frac{80}{\sqrt{160^2+80^2+120^2}} )=68.2^{o}$

Similarly angle between z axis and the vector is given by

$\theta_z =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{k}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{z\cdot k}{\sqrt{x^2+y^2+z^2}} )$

Applying values we get

$\theta_z=cos^{-1}(\frac{120}{\sqrt{160^2+80^2+120^2}} )=56.145^{o}$

5 0

4 years ago

engineering controls are devices such as self sheathing needles and sharps containers to block or eliminate the sharp risk. true

Paladinen [302]

Answer:

yes

Explanation:

The 1991 standard states, "engineering and work practice controls shall be used to eliminate or minimize employee exposure." The revision defines Engineering Controls as "controls (e.g., sharps disposal containers, self-sheathing needles, safer medical devices, such as sharps with engineered sharps injury protections ...

6 0

3 years ago

6.8.1: Function pass by pointer: Transforming coordinates. Define a function CoordTransform() that transforms its first two inpu

Black_prince [1.1K]

Answer:

The code is in the Explanation

Explanation:

#include <iostream>

using namespace std;

int main() {

int xValNew = 0;

int yValNew = 0;

CoordTransform(3, 4, xValNew, yValNew);

cout << "(3, 4) becomes " << "(" << xValNew << ", " << yValNew << ")" << endl;

return 0;}

6 0

3 years ago

Draw the logic circuit for each of the following. For each gate, determine if it generates either EVEN or ODD parity bit and fin

omeli [17]

Answer:

a) 4-input XOR, input data-1001 = 0 Even parity Bit

b) 5-input XOR, input data-10010 = 0 Even parity Bit

c) 6-input XOR, input data-101001 = 1 Even parity Bit

d) 7-input XOR, input data 1011011 = 1 Even parity Bit

Explanation:

a) 4-input XOR, input data-1001 ; generates 0 Even parity Bit

b) 5-input XOR, input data-10010 ; generates 0 Even parity Bit

c) 6-input XOR, input data-101001 ; generates 1 Even parity Bit

d) 7-input XOR, input data 1011011 ; generates 1 Even parity Bit

Attached below is the Logic circuits of the data inputs

8 0

3 years ago