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xz_007 [3.2K]
3 years ago
15

(TCO 4) In a CAN bus, there are three computers: Computer A, Computer B, and Computer C. The messages being sent are 2 milliseco

nds long. Computer A starts sending a message on the bus; 250 microseconds later, Computer B starts to send a message, and 250 microseconds after that, Computer C starts to send a message. When does Computer A retransmit its message?
A) After 2.5 milliseconds
B) After 1 millisecond
C) After 1.5 milliseconds
D) Never; Computer A will not retransmit.
Engineering
1 answer:
ladessa [460]3 years ago
4 0

Answer:

A) After 2.5 milliseconds

Explanation:

Given that :

In a CAN bus, there are three computers: Computer A, Computer B, and Computer C.

Length in time of the message sent are 2 milliseconds long

As the computer start sending the message; the message are being sent at a message duration rate of 2 milliseconds

250 microseconds later, Computer B starts to send a message; There is a delay of 250 microseconds = 0.25 milliseconds here at Computer B

and 250 microseconds after that, Computer C starts to send a message

Similarly; delay at Computer C = 0.25 milliseconds

Assuming T_{RT} is the retransmit time for Computer A to retransmit its message, Then :

T_{RT} = T_A + T_B + T_C

T_{RT} = 2 milliseconds + 0.25 milliseconds + 0.25 milliseconds

T_{RT} = 2.5 milliseconds

Thus; the correct option is A) After 2.5 milliseconds

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A heat engine operates between a source at 477°C and a sink at 27°C. If heat is supplied to the heat engine at a steady rate of
lara [203]

Answer:

T_C = 27+273.15 = 300.15 K

T_H = 477+273.15 = 750.15 K

And replacing in the Carnot efficiency we got:

e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%

W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}

Explanation:

For this case we can use the fact that the maximum thermal efficiency for a heat engine between two temperatures are given by the Carnot efficiency:

e = 1 -frac{T_C}{T_H}

We have on this case after convert the temperatures in kelvin this:

T_C = 27+273.15 = 300.15 K

T_H = 477+273.15 = 750.15 K

And replacing in the Carnot efficiency we got:

e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%

And the maximum power output on this case would be defined as:

W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}

Where Q_H represent the heat associated to the deposit with higher temperature.

4 0
3 years ago
Explicar el funcionamiento de un multímetro analógico.
Whitepunk [10]

Answer:

Un multímetro analógico funciona como un medidor de bobina móvil de imán permanente (PMMC) para tomar mediciones eléctricas

Explanation:

El multímetro analógico es un medidor o galvanómetro D'Arsonval que funciona según el principio de los medidores de bobina móvil de imán permanente (PMMC)

Un multímetro analógico está formado por un puntero de aguja unido a una bobina móvil colocada entre el polo norte y sur de un imán permanente dispuesto de tal manera que, cuando una corriente eléctrica fluye a través de la bobina, genera una fuerza de campo magnético que interactúa con el imán fuerza de campo de los imanes permanentes que hace que la bobina se mueva junto con el puntero de la aguja sobre un dial graduado

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3 years ago
How can you do this 5.2.4: Rating?
gizmo_the_mogwai [7]

Answer:

whats the question

Explanation:

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A plot of land is an irregular trangle with a base of 122 feet and a height of 47 feet what is the area of the plot?
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Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2
motikmotik

Answer:

The answer is "112.97 \ \frac{ft}{s}"

Explanation:

Air flowing into thep_1 = 20 \ \frac{lbf}{in^2}

Flow rate of the mass m  = 230.556 \frac{lbm}{s}

inlet temperature T_1 = 700^{\circ} F

PipelineA= 5 \times 4 \ ft

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}

      = \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}

V= \frac{mv}{A}

   = \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}

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