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Usimov [2.4K]
4 years ago
13

25 points!!

Engineering
1 answer:
adell [148]4 years ago
5 0
Answer 1:

Like most electrical mechanical devices, electric motors have mechanical bearings that eventually wear out.

The graphite brushes and commutator of older DC motors can also wear out over time

Answer2:

Question two is not specific enough!!
Are we talking DC?? Because if so, then yes, nothing will happen. Assuming you are insulted by ground.
Ac is a different story, as the two would light up dangerously and etc.

-your welcome!!
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If the open circuit voltage of a circuit containing ideal sources and resistors is measured at 6 V, while the current through th
Trava [24]

Given Information:

Current of Ideal source = Is = 100 mA = 0.100 A

Open circuit voltage = Voc = 6 V

Short circuit current = Isc = 200 mA = 0.200 A

Required Information:

Power absorbed by idea current source = ?

Answer:

Power absorbed by idea current source = 0.3 Watts

Explanation:

Ideal Current Source:

An ideal current source doesn't have internal resistance and provides constant current regardless of the voltage supplied to the circuit.

The power absorbed by the ideal source can be found using

P = Is²R

Where Is is the ideal source current and R can be found using

R = Voc/Isc

Where Voc is the open circuit voltage and Isc is the short circuit current.

R = 6/0.200

R = 30 Ω

Therefore, the power absorbed by the ideal current source is

P = Is²R

P = (0.100)²*30

P = 0.3 Watts

7 0
3 years ago
Which of the following describes beta testing?
Sergio039 [100]

Answer:

What is Beta Testing? Beta testing is an opportunity for real users to use a product in a production environment to uncover any bugs or issues before a general release. Beta testing is the final round of testing before releasing a product to a wide audience.

3 0
3 years ago
____ is based on the observation that the rate of increase in transistor density on microchips had increased steadily, roughly d
Ira Lisetskai [31]

Answer:

Moore's Law

Explanation:

An observation that the number of transistors in a dense integrated circuit doubles about every two years (24 months), was made by Gordon E. Moore, the co-founder of Intel, and this observation became Moore's Law in 1965.

Therefore, Moore's Law is based on the observation that the rate of increase in transistor density on microchips had increased steadily, roughly doubling every 18 to 24 months.

4 0
4 years ago
Paul is constructing an aquarium. The exterior of the aquarium in made of white marble. Which design principle element is being
sasho [114]

Answer:

B

Explanation:

because you can see in the text it is expressing that it is shape

7 0
4 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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