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Stels [109]
3 years ago
15

What is an obstacle that holds people back from getting their car fixed?

Engineering
2 answers:
docker41 [41]3 years ago
6 0

Answer:

Not having enough money to pay to get it fixed

Explanation:

Vlada [557]3 years ago
3 0

Answer:

1) Not having enough money

2) Not having the transportation to get to the car fixer place

3) Having nobody to watch the kids at home while you go get the cat fixed

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Fav Song Writer!!--Mine.... Alec Ben ,Tate Mcrae , Oliva Rodrigo , Why Dont We , Bille. THATS MINE...
MA_775_DIABLO [31]

Answer:

I like why don't we too, Prettymuch, Bts, Blackpink. Ateez, Staykids, etc...

Explanation:

7 0
3 years ago
Read 2 more answers
Consider a thin-walled cylindrical tube having a radius of 65 mm that is to be used to transport pressurized gas. (a) (10 points
AlladinOne [14]

Answer:

The minimum required thickness for a steel pressure vessel (t) is 2.275 mm

Explanation:

Minimum required thickness is the thickness of a material without corrosion allowance for each component  based on the appropriate design that consider pressure, mechanical and structural loading.

Given that:

radius (r) = 65 mm = 65 × 10⁻³ m

Factor of safety (N) = 3.5

Inside presssure (P_{in}) = 11 MPa

Outside pressure (P_{out}) = 1 MPa

Yield strength (\sigma_y) = 1000 MPa

Therefore:

\sigma_y =\frac{\sigma_y}{N}, Substituting values,

\sigma_y =\frac{\sigma_y}{N}=\frac{1000}{3.5}=285.714 MPa

The minimum required thickness for a steel pressure vessel (t) is given by the equation:

t=\frac{r.(P_{in}-P_{out})}{\sigma_y}. Substituting values

t=\frac{r.(P_{in}-P_{out})}{\sigma_y}=\frac{65*10^{-3} *(11-1)10^{6} }{285.714*10^{6} } =2.275*10^{-3} =2.275 mm

The minimum required thickness for a steel pressure vessel (t) is 2.275 mm

7 0
3 years ago
Input signal to a controller is​
alexgriva [62]

Answer:

were the cord plugs in

Explanation:

4 0
3 years ago
Hi, I have an assignment in which i needs to write a report on (Rationalization of electrical energy consumption) and i need cha
Zigmanuir [339]
^^^^^^^^^^^^^^^^^^^^^^^^^
6 0
3 years ago
An equilibrium mixture of 3 kmol of CO, 2.5 kmol of O2, and 8 kmol of N2 is heated to 2600 K at a pressure of 5 atm. Determine t
garri49 [273]

Answer:

x_{CO}=0.0203\\x_{O_2}=0.0926\\x_{CO_2}=0.227\\x_{N_2}=0.660

Explanation:

Hello,

In this case, we consider the reaction:

CO(g)+\frac{1}{2} O_2(g)\rightleftharpoons CO_2

For which the law of mass action is expressed as:

Kp=\frac{n_{CO_2}}{n_{CO}*n_{O_2}^{1/2}} (\frac{P}{n_{Tot}} )^{1-1-1/2}

Whereas the exponents are referred to the stoichiometric coefficients in the chemical reaction. Moreover, in table A-28 (Cengel's thermodynamics) the natural logarithm of the undergoing reaction at 2600 K is 2.801, thus:

K=exp(2.801)=16.46

In such a way, in terms of the change x the equilibrium goes:

16.46=\frac{x}{(3kmol-x)*(2.5kmol-0.5x)^{0.5}} (\frac{5}{13.5kmol-0.5x} )^{-0.5}

Hence, solving for x:

x=2.754kmol

Thus, the moles at equilibrium:

n_{CO}=3-2.754=0.246kmol\\n_{O_2}=2.5-0.5(2.754)=1.123kmol\\n_{CO_2}=x=2.754kmol\\n_{N_2}=8kmol

Finally the compositions:

x_{CO}=\frac{0.246}{0.246+1.123+2.754+8} =0.0203\\\\x_{O_2}=\frac{1.123}{0.246+1.123+2.754+8} =0.0926\\\\x_{CO_2}=\frac{2.754}{0.246+1.123+2.754+8} =0.227\\\\x_{N_2}=\frac{8}{0.246+1.123+2.754+8} =0.660

Best regards.

7 0
4 years ago
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