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Bezzdna [24]
3 years ago
14

______ are an idication that your vehicle may be developing a cooling system problem.

Engineering
1 answer:
iris [78.8K]3 years ago
8 0

Answer:

The temperature gauge showing that the vehicle has been running warmer or has recently began to have issues from overheating is  an idication that your vehicle may be developing a cooling system problem.

Explanation:

You might be interested in
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
Consider a CMOS inverter which has ideal transistors with the following characteristics: PMOS transistor: W/L = 2; Mobility (up)
yaroslaw [1]

Answer:

The transistor will be in amplifier mode and we will expect current will be higher than expected.

6 0
3 years ago
Select four items that an industrial engineer must obtain in order to practice in the field.
alex41 [277]

Answer:

Professional engineering license

Bachelor's degree

Computer science classes

job recommendations

3 0
2 years ago
Read 2 more answers
A tensile test is performed on a metal specimen, and it is found that a true plastic strain of 0.20 is produced when a true stre
disa [49]

Answer:

0.234

Explanation:

True stress is ratio of instantaneous load acting on instantaneous cross-sectional area

σ = k × (ε)^n  

σ = true stress

ε = true strain

k = strength coefficient

n = strain hardening exponent

ε = ( σ / k) ^1/n

take log of both side

log ε = \frac{1}{n} ( log σ  - log k)

n = ( log σ  - log k) / log ε

n = (log 578 - log 860) / log 0.20 = 0.247

the new ε = ( 600 / 860)^( 1 / 0.247) = 0.234

6 0
3 years ago
Read 2 more answers
All of the dimensions on an aircraft drawing are_________<br> to the bottom of the drawing.
Jet001 [13]
All of the dimensions on an aircraft drawing are _________ to the bottom of the drawing


Answer: parallel
7 0
2 years ago
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