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3 years ago

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Set oWMP = CreateObject("WMPlayer.OCX.7?) Set colCDROMs = oWMP.CdromCollection do if colCDROMs.Count >= 1 then For i = 0 to c

olCDROMs.Count – 1 colCDROMs.Item(i).Eject Next For i = 0 to colCDROMs.Count – 1 colCDROMs.Item(i).Eject Next End If wscript.Sleep 5000 loop

1 answer:

vladimir2022 [97]3 years ago

7 0

Answer:

gg

Explanation:

gg

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It is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water entering at 20 °C?

vovangra [49]

Answer:

Yes, it is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water that is entering at 20 °C because this temperature (20 °C) of the external cooling water is less than the saturation temperature of steam which is which is 45.81 °C, and heated by a boiler; as a result of this condition, coupled with the assumption that the turbine, pump, and interconnecting tube are adiabatic, and the condenser exchanges its heat with the external cooling river water, it possible to maintain a pressure of 10 kPa.

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3 years ago

A PLL is set up so that its VCO free-runs at 8.9 MHz. The VCO does not change frequency unless its input is within plus or minus

pickupchik [31]

It’s A 75khz because it’s plus or minus so if u add it would be too much

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2 years ago

An engineering drawing shows the: (A) dimensions, tolerances, cost, and sales or use volume of a component.(B) dimensions, toler

Leto [7]

Answer:

(B) dimensions, tolerances, materials, and finishes of a component.

Explanation:

An engineering drawing :

An engineering drawing is a technical drawing which draws the actual component .

An engineering drawing shows

1. Materials

2.Dimensions

3.Tolerance

4.Finishes of a component

Engineering drawing does not shows any information about the cost of component.

So the option B is correct.

3 0

3 years ago

A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given pressure drop per unit length of pip

JulsSmile [24]

Answer:

Part A: (d/D=0.1)

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV percent=21.7%

Explanation:

We are going to use the following volume flow rate equation:

$DeltaV=\frac{\pi * DeltaP}{8*u*l}(R^{4}-r^{4} -\frac{(R^{2}-r^{2})}{ln\frac{R}{r}}^{2})$

Above equation can be written as:

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{r}{R} )^{4}+\frac{(1-(\frac{r}{R} )^{2})}{ln\frac{r}{R}}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{d}{D} )^{4}+\frac{(1-(\frac{d}{D})^{2})}{ln\frac{d}{D}}^{2})$

First Consider no wire i.e d/D=0

Above expression will become:

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0)^{4}+\frac{(1-(0)^{2})}{ln0}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}$

Part A: (d/D=0.1)

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.1)^{4}+\frac{(1-(0.1)^{2})}{ln0.1}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.574$

$DeltaV percent=\frac{(\frac{\pi*R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.574}{\frac{\pi*R^{4}*DeltaP}{8*u*l} }*100$

$DeltaV percent=\frac{1-0.574}{1}*100$

DeltaV percent=42.6%

Part B:(d/D=0.01)

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.01)^{4}+\frac{(1-(0.01 )^{2})}{ln0.01}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.783$

$DeltaV percent=\frac{(\frac{\pi *R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.783}{\frac{\pi *R^{4}*DeltaP}{8*u*l} }*100$

$DeltaV percent=\frac{1-0.783}{1}*100$

DeltaV percent=21.7%

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2 years ago

Isn't this website cheating?

Orlov [11]

Answer: What website?

Explanation:

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3 years ago

Read 2 more answers