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3 years ago

One reason that driving lends itself to aggressive behavior is:

Engineering

2 answers:

Maurinko [17]3 years ago

5 0

Answer:

C. Drivers feel empowered by anonymity

Explanation:

Sorry for the late answer hope it helps you and others who need it.

STay Safe and healthy!

vlada-n [284]3 years ago

4 0

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Water enters a tank from two pipes, one with a flow rate of 0.3 kg/s and the other with a flow rate of 0.1 kg / s. The tank has

stiks02 [169]

Answer:

total amount of water after 2 min will be 84.4 kg/s

Explanation:

Given data:

one tank inflow = 0.1 kg/s

2nd tank inflow = 0.3 kg/s

3rd tank outflow = 0.03 kg/s

Total net inflow in tank is = 0.3 +0.1 =0.4 kg/s

From third point, outflow is 0.03 kg/s

Therefore, resultant in- flow = 0.4 - 0.03

Resultant inflow is = 0.37 kg/s

Tank has initially 40 kg water

In 2 min ( 2*60 sec), total inflow in tank is 0.37*60*2 = 44.4 kg

So, total amount of water after 2 min will be = 40+44.4 = 84.4 kg

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3 years ago

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: Dalton's Law of Partial Pressure 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 mole fraction:

$X_{i}$ = 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.

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3 years ago

Which of the following statements best describes the relationship between availability of new green building products and custom

Fed [463]

Answer:

Explanation:

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3 years ago

Select the correct answer.

andreev551 [17]

Answer:

Explanation:

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3 years ago

Read 2 more answers

1. For ball bearings, determine: (a) The factor by which the catalog rating (C10) must be increased, if the life of a bearing un

ElenaW [278]

Answer:

(b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.

Equation 11-1: F*L^(1/3) = Constant

Weibull parameters of example 11-3: xo = 0.02 (theta-xo) = 4.439 b = 1.483

Explanation:

(a)The Catalog rating(C)

Bearing life: $L_1 = L , L_2 = 2L$

Catalog rating: $C_1 = C , C_2 = ? ,$

From given equation bearing life equation,

$F\times\frac{1}{3} (L_1) = C_1 ...(1) \\\\ F\times\frac{1}{3} (L_2) =C_2...(2)$

we Dividing eqn (2) with (1)

$\frac{C_2}{C_1} =\frac{1}{3} (\frac{L_2}{L_1})\\\\ C_2 = C*(\frac{2L}{L})\frac{1}{3} \\\\ C_2 = 1.26 C$

The Catalog rating increased by factor of 1.26

(b) Reliability Increase from 0.9 to 0.99

$R_1 = 0.9 , R_2 = 0.99$

Now calculating life adjustment factor for both value of reliability from Weibull parametres

$a_1 = x_o + (\theta - x_o){ ln(\frac{1}{R_1} ) }^{\frac{1}{b}}$

$= 0.02 + 4.439{ ln(\frac{1}{0.9} ) }^{\frac{1}{1.483}} \\\\ = 0.02 + 4.439( 0.1044 )^{0.67}\\\\a_1 = 0.9968$

Similarly

$a_2 = x_o + (\theta - x_o){ ln(\frac{1}{R_2} ) }^{\frac{1}{b} }\\\\ = 0.02 + 4.439{ ln(1/0.99) }^{\frac{1}{1.483} }\\\\ = 0.02 + 4.439( 0.0099 )^{0.67}\\\\a_2 = 0.2215$

Now calculating bearing life for each value

$L_1 = a_1 * LL_1 = 0.9968LL_2 = a_2 * LL_2 = 0.2215L$

Now using given ball bearing life equation and dividing each other similar to previous problem

$\frac{C_2}{C_1} = (\frac{L_2}{L_1} )^{\frac{1}{3} }\\\\ C_2 = C* (\frac{0.2215L }{0.9968L} )^{1/3}\\\\ C_2 = 0.61 C$

Catalog rating increased by factor of 0.61

6 0

4 years ago