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LuckyWell [14K]
2 years ago
12

The specification limits are 49 /- 3. Assume that the data is normally distributed! estimate process capability ratio’s (cp, cpk

, pp, ppk). Is the process capable?
Physics
1 answer:
yarga [219]2 years ago
5 0

Estimate technique capability ratio (cp, cpk, pp, PPK) is Cp = 1.33 or greater. & Cpk price ≥ 1.33.

Cp, Cpk, Pp, and Ppk are all parameters (indices) that can help us to apprehend how our technique is working relative to the specs, or in different words, they degree how near our system is jogging to its specification limits. For necessities, we measure the process specs.

The Cp index is a fundamental indication of process capability. The Cp value is calculated using the specification limits and the same old deviation of the method. most agencies require that the method Cp = 1.33 or greater. Cp and Cpk, generally called technique functionality indices, are used to outline the ability of a technique to produce a product that meets necessities.

For stable tactics and normally distributed statistics, a Cpk price ≥ 1.33 should be executed. • For chronically unstable methods with output assembly specification and a predictable sample, a Ppk fee ≥ 1. sixty-seven have to be completed.”

Learn more about limits here:-brainly.com/question/27585271

#SPJ4

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A boat is heading due north across a river with a speed of 12.0 km/h relative to the water. The water in the river has a uniform
irga5000 [103]
<h2>The velocity of the boat relative to an observer standing on either bank = u = 18 \frac{km}{hr}</h2>

Explanation:

Let speed of the boat in still water = u \frac{km}{hr}

speed of the river water = v \frac{km}{hr}

Relative speed of the boat in the water against the river flow is given by

Upstream speed = u - v ------- (1)

⇒ u - v = 12 \frac{km}{hr} ------ (2)

Given that speed of the water = 6 \frac{km}{hr}

Now velocity of the boat is given From equation (2)

⇒ u = 12 + v

Put the value of v = 6 , we get

⇒ u = 12 + 6

⇒ u =  18 \frac{km}{hr}

therefore , the velocity of the boat relative to an observer standing on either bank = u = 18 \frac{km}{hr}

7 0
3 years ago
Automobile air bags use the decomposition of sodium azide as their sources of gas for rapid inflation, represented in the reacti
NeTakaya

Answer : The mass of NaN_3 required is 71.175 grams.

Explanation :

To calculate the moles of nitrogen gas, we use the equation given by ideal gas :

PV = nRT

where,

P = Pressure of nitrogen gas = 763 torr

V = Volume of the nitrogen gas = 40.0 L

n = number of moles of gas = ?

R = Gas constant = 62.364\text{ L.torr }mol^{-1}K^{-1}

T = Temperature of helium gas = 25^oC=273+25=298K

Putting values in above equation, we get:

763torr\times 40.0L=n\times 62.364\text{ L.torr }mol^{-1}K^{-1}\times 298K\\\\n=1.642mol

Now we have to calculate the moles of NaN_3.

The balanced chemical reaction is:

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

From the balanced chemical reaction, we conclude that

As, 3 moles of N_2 produced from 2 moles NaN_3

So, 1.642 moles of N_2 produced from \frac{2}{3}\times 1.642=1.095 moles NaN_3

Now we have to calculate the mass of NaN_3.

Molar mass of NaN_3 = 65 g/mol

\text{Mass of }NaN_3=\text{Moles of }NaN_3\times \text{Molar mass of }NaN_3

\text{Mass of }NaN_3=1.095mole\times 65g/mole=71.175g

Therefore, the mass of NaN_3 required is 71.175 grams.

5 0
4 years ago
A person on a daily diet of 2,500 calories should get no_more than <br> calories from fat each day.
ivann1987 [24]

Answer:

35%

Explanation:

3 0
3 years ago
Read 2 more answers
The index of refraction of a transparent substance is 1.7, and the speed of light in air is 3.00 · 108 s. Calculate the speed of
Natalija [7]
Given the index of refraction, n and speed of light in the vacuum, c, we can solve for the speed of light in the transparent substance by the equation below.
                                                n = c / v
where v is our unknown.
Substituting the known values,
                                            1.7 = (3 x 10^8 m/s) / v
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5 0
4 years ago
Consider a metal single crystal oriented such that the normal to the slip plane and slip direction are at angles of 60° and 35°,
Vedmedyk [2.9K]

Answer:

19.5324 MPa

Explanation:

Information provided

Angle between the normal to the slip plane with tensile axis, \alpha=60^{o}&#10;

Angle by slip direction with tensile axis, \beta=35^{o}

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Applied stress \sigma=12 MPa

Shear stress at slip plane

\tau=\sigma cos\alpha cos\beta

\tau=12cos60^{o}cos35^{o}=4.915 MPa

\tau hence crystal won’t yield

Applied stress, \sigma for crystal to yield is given by

\sigma=\frac {\tau_{c}}{cos\alpha cos\beta}

\sigma=\frac {8}{cos60cos35}=19.53239342 MPa

\sigma=19.5324 MPa&#10;

7 0
3 years ago
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