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2 years ago

When must an Assured Equipment Grounding Conductor Program (AEGCP) be in place?

Engineering

1 answer:

sasho [114]2 years ago

8 0

Answer:

Only when working around power lines

Explanation:

Assured Equipment Grounding Conductor Program (AEGCP) is applicable mainly in sites dealing with power. Therefore, the most appropriate worksite to have Assured Equipment Grounding Conductor Program is when working around power lines.

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The y component of velocity in a steady, incompressible flow field in the xy plane is v = -Bxy3, where B = 0.4 m-3 · s-1, and x

love history [14]

Answer:

attached below

Explanation:

3 0

2 years ago

For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about ____

7nadin3 [17]

Answer:

One

For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about _one___

<h3>what is supported mounted?</h3>

A structure that holds up or serves as a foundation for something else. Support is a synonym for mounting.

To learn more about it, refer

to brainly.com/question/25689052

#SPJ4

4 0

1 year ago

A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte

Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

$a(t)=5(1-\frac{t}{15})$

$a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})$

Integrating both sides we get

$\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c$

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

$\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D$

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

$x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m$

Thus the remaining 125 meters will be covered with a constant speed of

$v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s$

in time equalling $t_{2}=\frac{125}{37.5}=3.33seconds$

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0

2 years ago

A vector AP is rotated about the Z-axis by 60 degrees and is subsequently rotated about X-axis by 30 degrees. Give the rotation

Musya8 [376]

Answer:

R = $\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]$ $\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]$

Explanation:

The mappings always involve a translation and a rotation of the matrix. Therefore, the rotation matrix will be given by:

Let $\theta$ and $\alpha$ be the the angles 60⁰ and 30⁰ respectively

that is $\theta$ = 60⁰ and

$\alpha$ = 30⁰

The matrix is given by the following expression:

$\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]$ $\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]$

The angles can be evaluated and left in the surd form.

4 0

2 years ago

Name the seven physical qualities for which standards have been developed.

SIZIF [17.4K]

Answer:

HUMAN DEVELOPMENT

MOTOR BEHAVIOR

EXERCISE SCIENCE

MEASUREMENT AND EVALUATION

HISTORY AND PHILOSOPHY

UNIQUE ATTRIBUTES OF LEARNERS

CURRICULUM THEORY AND DEVELOPMENT

Explanation:

6 0

3 years ago