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3 years ago

11

A technician is troubleshooting a problem. The technician tests the theory and determines the theory is confirmed. Which of the

following should be the technician’s NEXT step?
A. Implement the solution.
B. Document lessons learned.
C. Establish a plan of action.
D. Verify full system functionality.

1 answer:

vladimir1956 [14]3 years ago

3 0

Answer:

b) Document lessons learned.

Explanation:

First he should do documentation

then C

then D

then A

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Find the gravitational potential energy of a light that has a mass of 13.0 kg and is 4.8 m above the ground

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The human circulation system has approximately 1×109 capillary vessels. Each vessel has a diameter of about 8 µm. Assuming cardi

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4 years ago

A simple generator is used to generate a peak output voltage of 19.0 V . The square armature consists of windings that are 6.65

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One of the efficient concepts that can help us find the number of turns of the cable is through the concept of induced voltage or electromotive force given by Faraday's law. The electromotive force or emf can be described as,

$\epsilon = NBA\omega$

Where,

N = Number of loops

B = Magnetic Field

A = Cross-sectional Area

$\omega$ = Angular velocity

Re-arrange to find N,

$N = \frac{\epsilon}{BA\omega}$

Our values are given as,

$\epsilon = 19V$

$B = 0.434T$

$\omega = 49.8\frac{rev}{s} (\frac{2\pi rad}{1 rev}) = 99.6\pi rad/s$

$A = (6.65*10^{-2})^2 m^2$

Replacing at our equation we have:

$N = \frac{\epsilon}{( 0.434)A\omega}$

$N = \frac{19}{( 0.434)((6.65*10^{-2})^2)(99.6\pi)}$

$N = 31.63 \approx 32$

Therefore the number of loops of wire should be wound on the square armature is 32 loops

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3 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

a

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

b

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

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3 years ago

A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that

$q_1+q_2=4.7$

and Force between charge particles is given by

$F=\frac{kq_1q_2}{r^2}$

$4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}$

$q_1\cdot q_2=0.522$

put the value of $q_1$

$q_2\left ( 4.7-q_2\right )=0.522$

$q_2^2-4.7q_2+0.522=0$

$q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}$

$q_2=0.114 C$

thus $q_1=4.586 C$

3 0

3 years ago