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Arte-miy333 [17]
3 years ago
13

Rests represent periods of __________ in a measure.

Engineering
1 answer:
DaniilM [7]3 years ago
8 0

Answer:

Rests represent periods of <em><u>silence</u></em> in a measure.

Explanation:

Hope this Helps!!!

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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%

5 0
2 years ago
A keyboard, monitor and mouse are all considered as computer hardware and a program is known as a software.
Iteru [2.4K]

Answer:

This should be true because program is a software and mouse and keyboard are specific computer hardware to make your exsperience with a computer more efficent and faster.

5 0
2 years ago
Reducing the CFM through the evaporator will cause the amount of moisture removed from the air to ____________________. A. stay
vodomira [7]

Answer:

C. increase

Explanation:

Pollution can be defined as the physical degradation or contamination of the environment through an emission of harmful, poisonous and toxic chemical substances.

On a related note, particulate pollution is a form of pollution that is responsible for the degradation of the environment.

Particulate matter is also referred to as particle pollution or atmospheric aerosol particles and it can be defined as a complex microscopic mixture of liquid droplets and solid particles that are suspended in air. Other forms of particle pollution includes space debris and marine debris. Some examples of particulate pollution are dusts, soot, dirt, smoke, etc.

Generally, when cubic feet-per-minute (CFM) are reduced through the use of an evaporator, it will cause the amount of moisture removed from the air to increase.

3 0
3 years ago
A microscope illuminator uses a transformer to step down the 120 V AC of the wall outlet to power a 12.0 V,50 W microscope bulb.
Anna35 [415]

Answer:2.88 ohms

Explanation:

R= V^2 / P

12^2/50

144/50

2.88 ohms

5 0
1 year ago
How do we define energy efficiency
marta [7]
Energy efficiency simply means using less energy to perform the same task – that is, eliminating energy waste. ... There are enormous opportunities for efficiency improvements in every sector of the economy, whether it is buildings, transportation, industry, or energy generation.
7 0
3 years ago
Read 2 more answers
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