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erica [24]
2 years ago
9

Develop a three-month moving average for this time series. Compute MSE and a forecast for month 8. If required, round your answe

rs to two decimal places. Do not round intermediate calculation.
Engineering
1 answer:
Trava [24]2 years ago
4 0

Answer:

The missing Table is;

Month  1  2  3  4  5  6  7

Value  24  13  21  14  20  23  15

MSE = 20.56

Forecast = 19.33

Explanation:

<u>Table Formation</u>

Month Value Forecast Error  Absolute Error Error Square

1          24    

2          13    

3          21    

4          14         19.33333 -5.33333 5.333333 28.44444

5          20           16                       4              4             16

6          23          18.33333 4.666667 4.666667 21.77778

7          15           19                   -4                       4             16

   

    20.55

    19.33

<u>Table Explanation</u>

In the above table, we first calculate the forecast column. A forecast is taken as the sum of the three entries in order and then taking out the average.

Error is calculated by subtracting the Actual value from the forecast value.

Then we take the absolute of the forecast value.

Then we take the square of the absolute of the forecasted values.

Now we calculate MSE as;

MSE = ( 28.44 + 16 + 21.78 + 16 ) / 4

<em>MSE = 20.56</em>

Forecast for 8th month  = ( 20 + 23 + 15 ) / 3

Forecast for 8th month  = 19.33

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The fatigue data for a brass alloy are given as follows: Stress Amplitude (MPa) Cycles to Failure 170 3.7 × 104 148 1.0 × 105 13
prohojiy [21]

Answer:

i) S–N plot is attached

ii) fatigue strength = 100 MPa

iii) fatigue life = 5.62 x 10^(5) cycles

Explanation:

i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)

ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.

So let's find the log of this and trace it on the graph attached.

Log(4 × 10^(6)) = 6.6

From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa

iii) The question says we should find the fatigue life for 120 MPa.

Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.

Thus,the fatigue life will be the inverse log of 5.75.

Thus, fatigue life = 10^(5.75)

Fatigue life = 5.62 x 10^(5)

8 0
2 years ago
what is the advantage of decreasing the field current of a separately excited dc motor below its nominal value
enyata [817]

Answer:

Ability to rotate at higher speeds

Explanation:

Constant K1 becomes greater than the other constant K2

This translates to that the motor being able to rotate at high speeds, without necessarily exceeding the nominal armature voltage.

The armature voltage is the voltage that is developed around the terminals of the armature winding of an Alternating Current, i.e AC or a Direct Current, i.e DC machine during the period in which it tries to generate power.

6 0
2 years ago
Instructions given by traffic police or construction flaggers _____. A. Are sometimes important to follow B. Are usually not imp
Anestetic [448]

Answer:

D. Overrule any other laws and traffic control devices.

Explanation:

Laws and traffic control devices are undoubtedly compulsory to be followed at every point in time to control traffic and other related situations. However, there are cases when certain instructions overrule these laws and traffic control devices. For example, when a traffic police is giving instructions, and though the traffic control devices too (such as traffic lights) are displaying their own preset lights to control some traffic, the instructions from the traffic police take more priority. This is because at that point in time, the instructions from the traffic control devices might not be just applicable or sufficient.

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6 0
2 years ago
Rigid bar ACB is supported by an elastic cir-cular strut DC having an outer diameter of 15 in. and inner diameter of 14.4 in. Th
Ne4ueva [31]

Answer:

The change in length of the circular strut DC = 0.0028 in.

The vertical displacement of the rigid bar at point B = 0.00378 in.

Explanation:

We have the following parameters or information in the question given above:

=> The outer diameter = 15 in., the inner diameter = 14.4 in., the modulus elasticity of E = 29,000 ksi, and the Point load P = 5kips.

The diagram showing the rigid bar ACB is supported by an elastic cir-cular strut DC  is given in  the attached picture below.

According to Newton's law of motion, it can be seen that the force on CD, that is FCD is equal and opposite to ACD. Hence, FCD = ACD.

Where FCD = p × [4 + 5] ÷ [sin Ф × 4].

kindly note that from the diagram sin Ф = 3/5, cos Ф = 4/5 and tan Ф = 3/4. Also p =5.

Hence, FCD =[ 5 × 9] ÷ [3/5 × 4] = 18.75 kip. So FCD = ACD.

The next thing here is to determine the area and length of CD, say the area of CD is G, thus, G = π/4 × [ 15² - 14.4²] = 13.854 in².

The lenght of CD is = √[4² + 3²] = √[16 + 9] = 5ft. Thus, 5 × 12 = 60in.

Hence, the change in length of the circular strut DC = [18.75 × 60] ÷ 13.854 × 29000 = 0.0028 in.

The vertical deflection of CD = 0.0028 × 3/5 = 0.00168 in.

We have that; 4 /CV = 9BV. Hence, BV = 9/4× CV.

(CV = vertical deflection of CD).

The vertical displacement of the rigid bar at point B = 9/ 4 × 0.00168 in = 0.00378 in.  

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