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Vitek1552 [10]
3 years ago
9

If we didn’t have the spark what could not happen?

Engineering
1 answer:
Sloan [31]3 years ago
8 0

Answer:

there's no electric current

You might be interested in
Optimum engine oil pressure at operating temperature and moderate engine load should be __________ ps
kobusy [5.1K]

Answer:

What Is Normal Oil Pressure Temperature? Oil temperatures typically range from 200 degrees to 200 degrees Fahrenheit, depending on the weather. The oil pressure ranges from 100psi to 10-15psi depending on the weather at start up.

Explanation:

3 0
2 years ago
1. A car engine transfers 3000J in 20 seconds. What is the power generated by the engine?
Scorpion4ik [409]

Answer:

  150 W

Explanation:

Power is energy per unit time. The power generated by the engine is ...

  (3000 J)/(20 s) = 150 J/s = 150 W

The engine generated 150 watts.

3 0
3 years ago
An ideal Carnot engine extracts of heat from a high-temperature reservoir at 1200 during each cycle, and rejects heat to a low-t
Brut [27]

Answer:

efficiency of the engine is 0.4167

Explanation:

given data

T high = 1200 K

T low = 700 K

solution

we get here efficiency that is express as

efficiency  = 1 - \frac{T\ low}{T \ high }    .........................1

put here value of both temperature we get efficiency  

efficiency  = 1 - \frac{700}{1200}

efficiency  = 1 - 0.58333

efficiency  = 0.4167

so efficiency of the engine is 0.4167

3 0
3 years ago
A large well-mixed tank of unknown volume, open to the atmosphere initially, contains pure water. The initial height of the solu
trasher [3.6K]

Answer:

The exact time when the sample was taken is = 0.4167337 hr

Explanation:

The diagram of a sketch of the tank is shown on the first uploaded image

Let A denote the  first inlet

Let B denote the second inlet

Let C denote the single outflow from the tank

From the question we are given that the diameter of A is = 1 cm = 0.01 m

                              Area of  A is  = \frac{\pi}{4}(0.01)^{2} m^{2}

                                                    = 7.85 *10^{-5}m^{2}

Velocity of liquid through A = 0.2 m/s

  The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

             = 0.2 *7.85*10^{-5} \frac{m^{3}}{s}

  The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              =  1039.8 * 0.2 * 7.85 *10^{-5} Kg/s

                              = 0.016324 \frac{Kg}{s}

From the question the diameter of B = 2 cm = 0.02 m

                                           Area of B = \frac{\pi}{4} * (0.02)^{2} m^{2} = 3.14 * 10^{-4}m^{2}

                                     Velocity of liquid through B = 0.01 m/s

The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

             = 3.14*10^{-4} *0.01 \frac{m^{3}}{s}

The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              = 1053 * 3.14*10^{-6} \frac{Kg}{s}

                              = 0.00330642 \frac{Kg}{s}

From the question The flow rate in term of volume of the outflow at the time of measurement is given as  = 0.5 L/s

And also from the question the mass of  potassium chloride  at the time of measurement is given as 13 g/L

So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              = 13\frac{g}{L} * 0.5 \frac{L}{s}

                              =  \frac{6.5}{1000}\frac{Kg}{s}       Note (1 Kg = 1000 g)

                              = 0.0065 kg/s

Considering potassium chloride

         Let denote the  rate at which liquid flows in terms of mass as   as \frac{dm}{dt} i.e change in mass with respect to time hence

           Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )

         

      (0.016324 + 0.00330642) - 0.0065 = \frac{dm}{dt}

          \int\limits {\frac{dm}{dt} } \, dx  =\int\limits {0.01313122} \, dx

      => 0.01313122 t = (m - m_{o})

  From the question  (m - m_{o})  is given as = 19.7 Kg

Hence the time when the sample was taken is given as

               0.01313122 t = 19.7 Kg

      =>  t = 1500.2414 sec

            t = .4167337 hours (1 hour = 3600 seconds)

5 0
4 years ago
What careers could you potential do if you
Margarita [4]

Answer:

Engineering careers. If you want to stay in engineering, your job opportunities are very much linked to your degree type, and you probably know what many of them are already. ...

Consulting. ...

Technical writing. ...

Business. ...

Investment banking. ...

Law. ...

Manufacturing and production. ...

Logistics and supply chain.

Explanation:

3 0
3 years ago
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