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Allisa [31]
1 year ago
12

A textile fiber manufacturer is investigating a new drapery yarn, which the company claims has a mean thread elongation of 12 ki

lograms with a standard deviation of 0. 5 kilograms. The company wishes to test the hypothesis H0:μ=12 against H1:μ<12 using a random sample of n=4 specimens. Calculate the P-value if the observed statistic is x⎯⎯=11. 25. Suppose that the distribution of the sample mean is approximately normal. Round your final answer to five decimal places (e. G. 98. 76543). Enter your answer in accordance to the question statement
Engineering
1 answer:
svet-max [94.6K]1 year ago
7 0

Answer:

The p-value of the hypothsis 0.00135

Explanation:

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A rigid tank contains 2 kg of N2 and 4 kg of Co2 at temperature of 25 C and 1 MPa. Find the partial pressure of each gas respect
lions [1.4K]

Answer: Partial pressures are 0.6 MPa for nitrogen gas and 0.4 MPa for carbon dioxide.

Explanation: <u>Dalton's</u> <u>Law</u> <u>of</u> <u>Partial</u> <u>Pressure</u> states when there is a mixture of gases the total pressure is the sum of the pressure of each individual gas:

P_{total} = P_{1}+P_{2}+...

The proportion of each individual gas in the total pressure is expressed in terms of <u>mole</u> <u>fraction</u>:

X_{i} = moles of a gas / total number moles of gas

The rigid tank has total pressure of 1MPa.

  • Nitrogen gas:

molar mass = 14g/mol

mass in the tank = 2000g

number of moles in the tank: n=\frac{2000}{14} = 142.85mols

  • Carbon Dioxide:

molar mass = 44g/mol

mass in the tank = 4000g

number of moles in the tank: n=\frac{4000}{44} = 90.91mols

Total number of moles: 142.85 + 90.91 = 233.76 mols

To calculate partial pressure:

P_{i}=P_{total}.X_{i}

For Nitrogen gas:

P_{N_{2}}=1.\frac{142.85}{233.76}

P_{N_{2}} = 0.6

For Carbon Dioxide:

P_{total}=P_{N_{2}}+P_{CO_{2}}

P_{CO_{2}} = P_{total}-P_{N_{2}}

P_{CO_{2}}=1-0.6

P_{CO_{2}}= 0.4

Partial pressures for N₂ and CO₂ in a rigid tank are 0.6MPa and 0.4MPa, respectively.

4 0
2 years ago
The water in a 25-m-deep reservoir is kept inside by a 140-m-wide wall whose cross section is an equilateral triangle as shown i
koban [17]

Answer:  (a) 9.00 Mega Newtons or 9.00 * 10^6 N

               (b)  17.1 m

Explanation:  The length of wall under the surface can be given by

                                            b=25m/sin(60)\\=28.867

The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.

F(resultant) = Pavg ( A) = (Patm +  \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N

Noting from the Bernoulli  equation that

Po/\rho g sin60 = 100000/1000 * 9.81* sin(60) = 11.77 m \\ \\

From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:

Yp = s+\frac{b}{2} +\frac{b^2}{s+\frac{b}{2}+Po/\rho g sin60}= 0+\frac{28.87}{2} +\frac{28.87^2}{0+\frac{28.87}{2}+100000 /1000 *9.81 sin60} = 17.1 m

Substituting the values gives us the the distance of the surface to be equal to = 17.1 m

7 0
3 years ago
A 2 m3 rigid tank initially contains air at 100 kPa and 22 degrees C. The tank is connected to a supply line through a valve. Ai
Vaselesa [24]

Answer:

a. 9.58kgs b. 340.32KJ

Explanation:

Volume of tank= 2m³

Initial Pressure Pi= 100KPa

Initial Temperature Ti= 22 C= 295K

Line Pressure P₁= 600 KPa

Line Temperature T= 22 C= 295K

Final Pressure P2= 600 KPa

Final Temperature T2= 77 C= 350K

Use Ideal Gas Equation

PV= mRT

P₁V₁= m₁RT₁

m₁= (100 x 2)/(0.287 x 295) = 2.3622kg

P₂V₂= m₂RT₂

m₂= (600 x 2)/(0.287 x 350) = 11.946 kg

Since valve is closed and no mass leave

m₁ + mi = m₂ + me

as per above condition me= 0

mi= m₂ - m₁ = 11.946 - 2.3622 = 9.5838kg

Applying energy equation

m₁u₁ + mihi + Q = m₂u₂ + mehe + W

me and W=0

m₁u₁ + mihi + Q = m₂u₂

m₁CvT₁ + miCpTi + Q =  m₂CvT₂

Q =   m₂CvT₂- m₁CvT₁ - miCpTi

Q = (11.946 x 0.717 350) - (2.3622 x 0.717 x 295) - (9.5868 x 1.004 x 295)

Q = -340.321 KJ (Negative sign doesn't matter as energy is not a vector quantity)

4 0
3 years ago
A(n) 78-hp compressor in a facility that operates at full load for 2500 h a year is powered by an electric motor that has an eff
Jet001 [13]

Answer: $17,206.13

Explanation:

Hi, to answer this question we have to apply the next formula:  

Annual electricity cost = (P x 0.746 x Ckwh x h) /η  

P = compressor power = 78 hp  

0.746 kw/hp= constant (conversion to kw)

Ckwh = Cost per kilowatt hour = $0.11/kWh  

h = operating hours per year = 2500 h  

η = efficiency = 93% = 0.93 (decimal form)  

Replacing with the values given :  

C = ( 78 hp x 0.746 kw/hp x 0.11 $/kwh x 2500 h ) / 0.93 = $17,206.13  

5 0
3 years ago
A flat site is being considered for a new school that will have a steel frame and brick façade. The steel columns will have a ma
ehidna [41]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

5 0
3 years ago
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