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algol13
3 years ago
7

For variables control, a circuit voltage will be measured using a sample of five circuits. The past average voltage for samples

of 5 units has been 3.4 volts, and the range has been 1.3 volts.
Required:
What would the upper and lower control limits be for the resulting control charts (average and range)?
Physics
1 answer:
astra-53 [7]3 years ago
8 0

Answer:

Average :

UCL = 4.15

LCL = 2.65

Range :

UCL = 2.75

LCL = 0

Explanation:

Given :

Sample size, n = 5

Average, X = 3.4

Range, R = 1.3

A2 for n = 5 ; equals 0.577 ( X chart table)

For the average :

Upper Control Limit (UCL) :

X + A2*R

3.4 + 0.577(1.3) = 4.1501

Lower Control Limit (LCL) :

X - A2*R

3.4 - 0.577(1.3) = 2.6499

FOR the range :

Upper Control Limit (UCL) :

UCL = D4*R

D4 for n = 5 ; equals = 2.114

UCL = 2.114*1.3 = 2.7482

Lower Control Limit (LCL) :

LCL = D3*R

D3 for n = 5 ; equals = 0

LCL = 0 * 1.3 = 0

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Answer:

0.25 m.

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We'll begin by calculating the spring constant of the spring.

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F = Ke

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K = 0.1/0.125

K = 0.8 N/m

Therefore, the force constant, K of spring is 0.8 N/m

Now, we can obtain the number in gap 1 in the diagram above as follow:

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F = Ke

0.2 = 0.8 x e

Divide both side by 0.8

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3 years ago
5g of ammonium nitrate was dissolved in 60g of water in an insulated container. The temperature at the start of the reaction was
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Energy absorbed by the reaction or energy lost by the water to the reaction,Q.

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Specific heat of water = c = 4.15 J\g ^oC

Change is temperature=\Delta T=19^oC-23^oC=-4^oC

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Answer:

5

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