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Airida [17]
3 years ago
5

Consider a system with two tasks, Task1 and Task2. Task1 has a period of 200 ms, and Task2 has a period of 300 ms. All tasks ini

tially tick at time 0. 1. How many times will Task1 have ticked after 1000 ms? 2. How many times will Task2 have ticked after 1000 ms? 3. After time 0, when do both Task1 and Task2 next tick at the same time? 4. What is the largest value for timerPeriod that allows both
Engineering
1 answer:
Murrr4er [49]3 years ago
4 0

<u>Explanation:</u>

Task 1 time period = 200ms, Task 2 time period = 300ms

Task ticked = \frac{1000ms}{200ms}= 5  →  5 times

Task 2 ticked =\frac{1000ms}{300ms} = 3.33 → 3 times

At 600 ms → 200ms 200ms 200ms

                     300ms → \frac{30ms}{60ms}

Largest time period = H.C.M of (200ms, 300ms)

                                 = 600ms

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D. Inspecting the muffler

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3 years ago
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A school is playing $0.XY per kWh for electric power. To reduce its power bill, the school installs a wind turbine with a rated
Semmy [17]

Answer: Your question has some missing figures so kindly plug in the values into the solution provided to get the exact amount of money saved

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<u>Calculating  the amount of electric power generated by wind turbine</u>

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Why is the lubrication system of an internal combustion engine equipped with an oil filter?
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Consider a 2-shell-passes and 8-tube-passes shell-and-tube heat exchanger. What is the primary reason for using many tube passes
Maru [420]

Answer:

See explanation

Explanation:

Solution:-

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5 0
3 years ago
) A shaft encoder is to be used with a 50 mm radius tracking wheel to monitor linear displacement. If the encoder produces 256 p
andrey2020 [161]

Answer:

number of pulses produced =  162 pulses

Explanation:

give data

radius = 50 mm

encoder produces = 256 pulses per revolution

linear displacement = 200 mm

solution

first we consider here roll shaft encoder on the flat surface without any slipping

we get here now circumference that is

circumference = 2 π r .........1

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so now we get number of pulses produced

number of pulses produced = \frac{linear\ displacement}{circumference} × No of pulses per revolution .................2

number of pulses produced = \frac{200}{314.16} × 256

number of pulses produced =  162 pulses

5 0
3 years ago
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