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3 years ago

6

While discussing armature testing: Technician A says to test for shorts, place the ohmmeter leads on a commutator segment and ar

mature shaft. Technician B says that continuity should exist between the commutator and the armature shaft. Who is correct?

1 answer:

KIM [24]3 years ago

7 0

Answer:

B says that continuity should exist between the commutator and the armature shaft.

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36. What is mass? (4.4)

KatRina [158]

Answer:

Mass, in physics, quantitative measure of inertia, a fundamental property of all matter.

Explanation:

Mass is the matter that makes up objects

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3 years ago

Has anyone lost faith in humanity ✌️

gregori [183]

‼️‼️‼️thank u for the freebie lolllll shdisndidneidn

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3 years ago

As a project manager of Permagam Construction you want to plan renting a fleet of 25 cu yd tractor-scrapers and have them haulin

dlinn [17]

Answer:

See explaination

Explanation:

Rolling resistance which in some occassions can be called rolling friction or rolling drag, is the force resisting the motion when a body rolls on a surface.

In order to calculate our rolling resistance, there should be a force.

Please kindly check attachment for the step by step solution of the given problem.

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3 years ago

An equilibrium mixture of 3 kmol of CO, 2.5 kmol of O2, and 8 kmol of N2 is heated to 2600 K at a pressure of 5 atm. Determine t

garri49 [273]

Answer:

$x_{CO}=0.0203\\x_{O_2}=0.0926\\x_{CO_2}=0.227\\x_{N_2}=0.660$

Explanation:

Hello,

In this case, we consider the reaction:

$CO(g)+\frac{1}{2} O_2(g)\rightleftharpoons CO_2$

For which the law of mass action is expressed as:

$Kp=\frac{n_{CO_2}}{n_{CO}*n_{O_2}^{1/2}} (\frac{P}{n_{Tot}} )^{1-1-1/2}$

Whereas the exponents are referred to the stoichiometric coefficients in the chemical reaction. Moreover, in table A-28 (Cengel's thermodynamics) the natural logarithm of the undergoing reaction at 2600 K is 2.801, thus:

$K=exp(2.801)=16.46$

In such a way, in terms of the change $x$ the equilibrium goes:

$16.46=\frac{x}{(3kmol-x)*(2.5kmol-0.5x)^{0.5}} (\frac{5}{13.5kmol-0.5x} )^{-0.5}$

Hence, solving for $x$ :

$x=2.754kmol$

Thus, the moles at equilibrium:

$n_{CO}=3-2.754=0.246kmol\\n_{O_2}=2.5-0.5(2.754)=1.123kmol\\n_{CO_2}=x=2.754kmol\\n_{N_2}=8kmol$

Finally the compositions:

$x_{CO}=\frac{0.246}{0.246+1.123+2.754+8} =0.0203\\\\x_{O_2}=\frac{1.123}{0.246+1.123+2.754+8} =0.0926\\\\x_{CO_2}=\frac{2.754}{0.246+1.123+2.754+8} =0.227\\\\x_{N_2}=\frac{8}{0.246+1.123+2.754+8} =0.660$

Best regards.

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4 years ago

assume a five layer network model. There are 700 bytes of application data. There is a 20 bye header at the transport layer, a 2

amm1812

Answer: The overhead percentage is 7.7%.

Explanation:

We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.

We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.

So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:

OH % = (58 / 758) * 100 = 7.7 %

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3 years ago