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LekaFEV [45]
3 years ago
15

Which type of memory is programmed at the factory? RAM ROM Cache or Virtual memory

Engineering
1 answer:
12345 [234]3 years ago
5 0

Answer:

RAM, which stands for random access memory, and ROM, which stands for read-only memory, are both present in your computer. RAM is volatile memory that temporarily stores the files you are working on. ROM is non-volatile memory that permanently stores instructions for your computer.

Explanation:

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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
Thoughts about drinking and driving
poizon [28]

Answer: it’s very bad behavior and can be dangerous

Explanation:

8 0
3 years ago
Read 2 more answers
We would like to measure the density (p) of an ideal gas. We know the ideal gas law provides p= , where P represents pressure, R
Nostrana [21]

Answer: =

Explanation:

=    P / (R * T) P- Pressure, R=287.058, T- temperature

From the given that

Sample mean(pressure) = 120300 Pa

Standard deviation (pressure) = 6600 Pa

Sample mean(temperature) = 340K

Standard deviation(temperature) = 8K

To calculate the Density;

Maximum pressure = Sample mean(pressure) + standard deviation (pressure) = 120300+6600 = 126900 Pa

Minimum pressure = Sample mean (pressure) - standard deviation (pressure)= 120300-6600 = 113700 Pa

Maximum temperature = Sample mean (temperature) + standard deviation (temperature) = 340+8 = 348K

Minimum temperature = Sample mean (temperature) - standerd deviation (temperature) = 340-8 = 332K

So now to calculate the density:

Maximum Density= Pressure (max)/(R*Temperature (min))= 126900/(287.058*332)= 1.331

Minimum density=Pressure(min)/(R*Temperature (max))= 113700/(287.058*348)= 1.138

Average density= (density (max)+ density (min))/2= (1.331+1.138)/2= 1.2345

cheers i hope this helps

5 0
3 years ago
In a CNC machining operation, the has to be moved from point (5, 4) to point(7, 2)along a circular path with center at (7,2). Be
notka56 [123]

Answer: hello your question is incomplete below is the complete question

answer:

N010 GO2 X7.0 Y2.0 15.0 J2.0  ( option 1 )

Explanation:

Given that the NC machining has to be moved from point ( 5,4 ) to point ( 7,2 ) along a circular path

GO2 = circular interpolation in a clockwise path

G91 = incremental dimension

<em>hence the correct option is </em>:

N010 GO2 X7.0 Y2.0 15.0 J2.0  

6 0
3 years ago
Maintain a distance of at least
Virty [35]
The answer is c. 4 seconds
8 0
3 years ago
Read 2 more answers
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