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Lerok [7]
3 years ago
11

X+3=2x=??No spamming​

Engineering
1 answer:
PtichkaEL [24]3 years ago
4 0

Answer:

x+3=2

x=2-3꧁

꧁꧁꧂꧂꧂꧂

x=-1

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The light colored material in areas of the mood is the earliest crust on the moon.
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The components of your sample mixture are present in trace amounts. Assuming that you will be using GC, what kind of sample intr
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Explanation:

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In a parallel circuit, as more resistances are added, what happens to the total circuit resistance?
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The flow of the electrons gets slower the more resistance added

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2 years ago
Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh
Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

R_a=e^{-\left ( \frac{T-r}{n}\right )B}

R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}

R_a=e^{-4.4}=0.01227

For B

R_b=e^{-\left ( \frac{300-336}{100}\right )3}

R_b=e^{1.08}=2.944

B is better for 100 hours

(b)For 750 hours

R_a=e^{-0.5866}=0.55621

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3 0
3 years ago
The elementary liquid-phase series reaction
liraira [26]

Answer:

Concentration of A: \frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }

Concentration of B: \frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1}  } (e^{-k_{1}t } -e^{-k_{2}t } )

Concentration of C: \frac{C_{C} }{C_{Ao} } =1+\frac{k_{1} }{k_{2}-k_{1}  } e^{-k_{2}t } -\frac{k_{2} }{k_{2}-k_{1}  } e^{-k_{1} t}

the image shows the graphs of the three concentrations

Explanation:

We have the reaction:

A ------->k1--------->B------------->k2--------->C

Each reaction:

r_{A} =-k_{1} C_{A} \\r_{B} =k_{1} C_{A} -k_{2} C_{B} \\r_{C} =k_{2} C_{C}

Where Cn is the concentration of each specie (A,B,C)

The mass balance for A:

-\frac{dC_{A} }{dt} =-r_{A} \\-\frac{dC_{A} }{dt}=k_{1} C_{A} \\-\int\limits^y_x {\frac{dC_{A} }{dt} } \,=k_{1} t\\\frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }

Where x=CAo and y=CA

The mass balance for B:

-\frac{dC_{B} }{dt} =-r_{B} \\-\frac{dC_{B} }{dt}=k_{2} C_{B} -k_{1} C_{A} \\\frac{dC_{B} }{dt}+k_{2} C_{B}=k_{1} C_{A}\\\frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1}  } (e^{-k_{1}t }-ex^{-k_{2}t }  )

The mass balance for C:

\frac{C_{C} }{C_{Ao} } =1-\frac{C_{A} }{C_{Ao} } -\frac{C_{B} }{C_{Ao} } \\\frac{C_{C} }{C_{Ao} }=1+\frac{k_{1} }{k_{2}-k_{1}  } e^{-k_{2} t}-\frac{k_{2} }{k_{2}-k_{1}  }  e^{-k_{1}t }

The maximum concentration of C is:

C_{Cmax} =C_{Ao} (\frac{k_{2} }{k_{1} } )^{\frac{k_{2} }{k_{2}-k_{1}  }}  =1.6(\frac{0.01}{0.4} )^{\frac{0.01}{0.01-0.4} } =1.76mol/dm^{3}

and the maximum time is:

t_{max} =\frac{ln\frac{k_{2} }{k_{1} } }{k_{2}-k_{1}  } =\frac{ln\frac{0.01}{0.4} }{0.01-0.4} =9.4 h

6 0
3 years ago
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