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3 years ago

What is engineering?

Engineering

1 answer:

ANTONII [103]3 years ago

8 0

Answer:

the branch of science and technology concerned with the design, building, and use of engines, machines, and structures. Anything that involves engines, wires, etc. basically

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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

if this light stays on after the initial startup of the vehicle it signals a warning that there is something wrong with the serv

NemiM [27]

Service brake system indicator is the warning that there is something wrong with the service brake system. Hence option f is correct.

<h3>What is indicator?</h3>

Amber-colored indicator lights can be found at the front, back, and occasionally on the left and right sides of the vehicle. Whether you're turning left, right, or into oncoming traffic, you use your indicators to signal your planned change of direction.

When this light turns on, one of two things will happen. Either the parking brake is engaged or the hydraulic fluid (brake fluid) in the master cylinder is low. Your brakes are made up of a system of hydraulic oil-filled tubes called brake lines.

Thus, service brake system indicator is the warning that there is something wrong with the service brake system. Hence option f is correct.

To learn more about indicator, refer to the link below:

brainly.com/question/28093573

#SPJ1

7 0

1 year ago

What is the name of the part that supports the headlight assembly?

Shkiper50 [21]

A headlight assembly is made up of many parts that work together smoothly. One of these components is the headlight bracket. Sometimes known as “headlight mounting bracket,” it keeps the headlamp affixed to the header. The headlight bracket is made from sturdy material like plastic or steel.

8 0

3 years ago

4.71 A full-wave rectifier circuit with a 1-kΩ operates from a 120-V (rms) 60-Hz household supply through a 6-to-1 transformer h

brilliants [131]

Answer:

$V_{p (load)} = 28,3 V - 0,7 V = 27,6 V$

$V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V$

The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.

$I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA$

Explanation:

The peak voltage after the 6 to 1 step down is $V_{p} = \frac{120}{6} \sqrt{2} = 28,3V$ . Then, the peak voltage of the rectified output is $V_{p} - [tex]V_{avg} = \frac{2V_{p (load)} }{\pi} = \frac{55,2 V}{\pi } = 17,6 V$ V_{d}[/tex] and according to the statement, the diodes can be modeled to be $V_{d} = 0,7 V$ . Then, the peak voltage in the load is $V_{p (load)} = 28,3 V - 0,7 V = 27,6 V$ .

The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.

The average output voltage is calculated as:

$V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V$

The average current in the load is calculated as:

$I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA$

4 0

4 years ago

Factors of production examples for a home?

maks197457 [2]

Answer:

The four factors of production are land, labor, capital, and entrepreneurship. They are the inputs needed for supply. They produce all the goods and services in an economy. That's measured by gross domestic product.

please give brainlist

hope this helped........

6 0

4 years ago