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Shalnov [3]
3 years ago
7

Estimate the uncertainty in a 22 m/sec air velocity measurement using a Pitot tube at 20C. Assume the atmospheric pressure is 1

00 kPa and that R and the density of water, H O2  are known with high enough accuracy to consider them as constants (zero uncertainty). The pressure for each tap is measured with separate manometers that have a resolution of 0.5 mm. The manometers use water as the fluid and each are oriented vertically. The absolute pressure (which gives you the density of air) is known with an uncertainty of 0.1 kPa. The temperature is known with an uncertainty of 1C. Assume the stagnation pressure, pt, is indicated by a static height of 7 mm.

Engineering
1 answer:
ivanzaharov [21]3 years ago
4 0

Answer:

Check the explanation

Explanation:

In calculating the second version of velocity that is expected velocity, which involves the division of the overall amount of estimated story points by the amount of sprints. Take for instance, if the development team estimates a total of 150 points over five sprints, then we can say that the team's expected velocity would be 30 points per sprint.

Kindly check the attached image below to see the step by step explanation to the question above.

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Answer:

a.

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(i) Memory-less - It is not memory-less because the given system is depend on past or future values.

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b.

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(i) Memory-less - It is memory-less because the given system is not depend on past or future values.

(ii) Causal - It is casual because the present value of output does not depend on the future value of input.

(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .

For example - for x[n] = 0 , y[n] = cos(0) = 1

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(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.

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