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3 years ago

When troubleshooting an inoperative power window motor, first check the:

Engineering

1 answer:

Gwar [14]3 years ago

3 0

When troubleshooting an inoperative power window motor, first check the:

fuse for the appropriate circuit. wiring to the motor. switch for the motor ground for the appropriate circuit will be work in the following way.

Explanation:

Use a test light to check the voltage at the motor of the window that is inoperative.

Press the power window switch during the testing process. If there is no voltage at the motor, there may be problems in the wiring that runs from the motor to the switch.
Check for short, loose or open wires.
Causes of power window malfunctions -Window malfunctions are typically caused from either a faulty window regulator (also called a window track), or a broken motor, cable pulley or window switch. Intermittent problems can cause windows to stop working temporarily only to work again and have more problems later.
Symptoms of a Bad or Failing Window Motor / Regulator Assembly. Common signs include having to press multiple times to roll the window up or down, slower or faster window speed, and clicking sounds from the door.
What to Do When Your Power Window Won't Go Up -1.Remove the door panel. 2.Disconnect the window from the motor. 3.Access and disengage the motor. 4.Reconnect the window to the motor and raise. 5.Replace the door panel.
Once your window is all the way down, hold the button down for 2-5 seconds. Release the button after you've held it down for a short period of time. On some vehicles, the required time to reset the window is 2 seconds.

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3 years ago

A two-dimensional flow field described by

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Answer:

the answer is

Explanation:

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3 years ago

Suppose we use radix sort to sort the English-language strings below using standard lexicographic ordering (i.e. sort in alphabe

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Answer:

a) 4 passes are required to sort the string.

b) 4

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iii) PART

iv) TARP

v) TARP

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O(d*(n+10)), n is no. of integers

Explanation:

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3 years ago

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Answer:

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Explanation:

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3 years ago

For the speed equation along centerline of a diffuser, calculate the fluid acceleration along the diffuser centerline as a funct

Marrrta [24]

Answer:

$a = v\cdot \frac{dv}{dx}$ , $v (x) = v_{in}\cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x \right]^{-1}$ , $\frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1 \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}$

Explanation:

Let suppose that fluid is incompressible and diffuser works at steady state. A diffuser reduces velocity at the expense of pressure, which can be modelled by using the Principle of Mass Conservation:

$\dot m_{in} - \dot m_{out} = 0$

$\dot m_{in} = \dot m_{out}$

$\dot V_{in} = \dot V_{out}$

$v_{in} \cdot A_{in} = v_{out}\cdot A_{out}$

The following relation are found:

$\frac{v_{out}}{v_{in}} = \frac{A_{in}}{A_{out}}$

The new relationship is determined by means of linear interpolation:

$A (x) = A_{in} +\frac{A_{out}-A_{in}}{L}\cdot x$

$\frac{A(x)}{A_{in}} = 1 + \left(\frac{1}{L}\right)\cdot \left( \frac{A_{out}}{A_{in}}-1\right)\cdot x$

After some algebraic manipulation, the following for the velocity as a function of position is obtained hereafter:

$\frac{v_{in}}{v(x)} = 1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1\right) \cdot x$

$v(x) = \frac{v_{in}}{1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x}$

$v (x) = v_{in}\cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x \right]^{-1}$

The acceleration can be calculated by using the following derivative:

$a = v\cdot \frac{dv}{dx}$

The derivative of the velocity in terms of position is:

$\frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1 \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}$

The expression for acceleration is derived by replacing each variable and simplifying the resultant formula.

8 0

3 years ago

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