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SIZIF [17.4K]
3 years ago
10

What sub-discipline of Mechanical Engineering focuses

Engineering
1 answer:
Komok [63]3 years ago
8 0
Automotive I believe
You might be interested in
Column arrays: Transpose a row array Construct a row array countValues with elements 1 to endValue, using the double colon opera
White raven [17]

Answer:

Matlab code with step by step explanation and output results are given below

Explanation:

We have to construct a Matlab function that creates a row vector "countValues" with elements 1 to endValue. That means it starts from 1 and ends at the value provided by the user (endValue).  

function countValues = CreateArray(endValue)

% Here we construct a row vector countValues from 1:endValue

     countValues = 1:endValue;

% then we transpose this row vector into column vector

     countValues = countValues';

 end

Output:

Calling this function with the endValue=11 returns following output

CreateArray(11)

ans =

    1

    2

    3

    4

    5

    6

    7

    8

    9

   10

   11

Hence the function works correctly. It creates a row vector then transposes it and makes it a column vector.

7 0
3 years ago
How do I draw this from the side?
tangare [24]

Answer:

draw it 3D

Explanation:

because it's a 3D picture

6 0
3 years ago
What pounds per square inch is required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water?
Brut [27]
2.31 ft/psig

1.98 psig would equalize feet of head 4’7”

4.583/2.31 = 1.98. .583 is 7”/12”


So round off to 2.0 psig to get steady stream of bubbles
4 0
3 years ago
A classroom which normally contains 26 people is to be air- conditioned. A person in the classroom typically dissipates heat at
lutik1710 [3]

Answer:

Explanation:

Heat dissipated by 26 people per second

= 26 x 360 x 1000 / (60 x 60)

= 2600 J

Heat dissipated by bulbs per second

= 15 x 40

= 600 J

Heat dissipated by fans per second

200 J

Heat dissipated by air coming in

= 200 x 10 x 1000 / (60 x 60)

=555.55 J

Heat gain by walls

= 5000 x 1000 / (60 x 60)

= 1388.88 J

Total heat gain per second

= 2600 + 600+200+555.55

= 3955.55 J

Capacity of air-conditioner required

= 3955.55 J/s

= 3.9 kJ/s

= 3.9kW

= 4 kW

7 0
3 years ago
The phasor technique makes it pretty easy to combine several sinusoidal functions into a single sinusoidal expression without us
devlian [24]

Answer:

The phasor technique can't be applied directly in the following cases:

a) 45 sin(2500t – 50°) + 20 cos(1500t +20°)

b) 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°)

c)  -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)

d)  75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)

Explanation:

For a) and c), it is not possible to use the phasor technique, due this technique is only possible when the sinusoidal signals to be combined are all of the same frequency.

This is due to the vector representing a signal is showed as a fixed vector in the graph( which magnitude is equal to the amplitude of the sinusoid and his angle is the phase angle with respect to cos (ωt)), which is rotating at an angular speed equal to the angular frequency of the sinusoidal signal that represents, like a radius that shows a point rotating in a circular uniform movement.

This rotating vector represents a sinusoidal signal, in the form of a cosine (as the real part of the complex function e^{j(wt+\alpha)}), so it is not possible to combine with functions expressed as a sine, even though both  have  the same frequency.

If we look at the graphs of cos (ωt) and sin (ωt) we can say that the sin lags the cos in 90º, so we can say the following:

sin (ωt) = cos (ωt-90º)

This means that in order to be able to represent a sine function  as a cosine, we need to rotate it 90º in the plane clockwise.

This is the reason why before doing this transformation, it is not possible to use the phasor technique for b) and d).

8 0
3 years ago
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