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konstantin123 [22]

2 years ago

12

Technician A says that a way to prevent galvanic corrosion is to duplicate the original installation method. Technician B says t

hat a way to prevent galvanic corrosion is to reuse coated bolts. Who is right

1 answer:

sertanlavr [38]2 years ago

8 0

Answer:

Technician A

Explanation:

Galvanic corrosion is not on only one metal alone but caused when two metals are interacting. Thus, Duplicating the original installation method is a better option because re-using a coated bolt doesn't prevent galvanic corrosion because both materials must be coated and not just the bolt and in technician B's case he is coating just the bolt. Thus, technician B's method will not achieve prevention of galvanic corrosion but technician A's method will achieve it.

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Part A - Transmitted power A solid circular rod is used to transmit power from a motor to a machine. The diameter of the rod is

uranmaximum [27]

Answer:

Explanation:

Given that solid circular rod rotates at constant speed and neglecting losses throughout the system, power is calculated as the product of torque and angular speed. That is to say:

$\dot W = T \cdot \omega$

There is a formula that relates torque with shear stress:

$\tau = \frac{T \cdot D}{2 \cdot J}$

Where $J$ is the torsion module, whose formula for a solid circular cross section is:

$J = \frac{\pi \cdot D^{4}}{32}$

The tension module is calculated herein:

$J = 3.835 in^{4}$

Maximum allowed torsion is found by isolating it from shear stress equation:

$T_{max} = \frac{2 \cdot J \cdot \tau_{max}}{D}$

$T_{max} = 31.907 kip \cdot in\\T_{max} = 2.659 kip \cdot ft$

Then, maximum transmissible power is determined directly:

$\dot W = (2.659 kip \cdot ft) \cdot (2)\cdot (\pi) \cdot (170 rpm)\\\dot W \approx 2840.188 \frac{kip \cdot ft}{min} \\\dot W \approx 86.066 h.p.$

8 0

3 years ago

Electricians will sometimes call ______ "disconnects" or a "disconnecting means."

Natali5045456 [20]

I want to say D but i’m not 100% sure

5 0

2 years ago

Nec ________ covers selection of time-delay fuses for motor- overload protection.

Murljashka [212]

Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

<h3>What article in the NEC covers motor overloads?</h3>

Article 430 that is found in National Electrical Code (NEC) is known to be state as “Motors, Motor Circuits and Controllers.” .

Note that the article tells that it covers areas such as motors, motor branch-circuit as well as feeder conductors, motor branch-circuit and others.

Therefore, Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

Learn more about motor- overload from

brainly.com/question/20738481

#SPJ1

6 0

1 year ago

Milk has a density of as much as 64.6 lb/ft3. What is the gage pressure at the bottom of the straw 6.1 inches deep in the milk?

gregori [183]

Answer:

Explanation:

1 inch is 0.0833333feet

6.1 inches is 0.5083 feet

Density = mass/volume

64.6 = mass/0.50833

mass = 64.6 x 0.5083 =32.83618lb

3 0

3 years ago

Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,

Lyrx [107]

Answer:

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$

Explanation:

$v_0$ = Initial velocity of ball A

$v_A=v_0\cos45^{\circ}$

$v_B$ = Initial velocity of ball B = 0

$(v_A)_n'$ = Final velocity of ball A

$v_B'$ = Final velocity of ball B

$e$ = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

$mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

Coefficient of restitution is given by

$e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

$(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

$v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

Adding the above two equations we get

$2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0$

$\boldsymbol{\therefore v_B'=0.6364v_0}$

$(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0$

From the conservation of momentum along the plane of contact we have

$(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}$

$v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}$

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$ .

5 0

3 years ago