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Arte-miy333 [17]

3 years ago

13

Rests represent periods of __________ in a measure.

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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Which branch of engineering most closely relates to mechanical engineering?

Rus_ich [418]

Answer:

C

Explanation:

I COULD be wrong, i'm not sure but im confident its c

4 0

3 years ago

Tech A says that in some cases, the electronic brake control module can be programmed with a new tire size to restore proper ele

Vlad [161]

Answer:

Both Techs A and B

Explanation:

Electronic braking systems are controlled by the electronic brake control module. It is a microprocessor that processes information from wheel-speed sensors and the hydraulic brake system to determine when to release braking pressure at a wheel that's about to lock up and start skidding and activates the anti lock braking system or traction system when it detects it is necessary.

Some electronic brake control modules can be programmed to the size of the vehicle's new tires to restore proper electronic brake control performance. While others may require replacing the module to match the module's programming to the installed tire size. So, both technicians A and B are correct.

3 0

3 years ago

1.0•10^-10 standard form

Drupady [299]

Answer:

$1.0 * 10^{-10} = 0.0000000001$

Explanation:

Given

$1.0 * 10^{-10}$

Required

Convert to standard form

$1.0 * 10^{-10}$

From laws of indices

$a^{-x} = \frac{1}{a^x}$

So, $1.0 * 10^{-10}$ is equivalent to

$1.0 * 10^{-10} = 1.0 * \frac{1}{10^{10}}$

$1.0 * 10^{-10} = 1.0 * \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}$

$1.0 * 10^{-10} = 1.0 * \frac{1}{10000000000}$

$1.0 * 10^{-10} = 1.0 * 0.0000000001$

$1.0 * 10^{-10} = 0.0000000001$

Hence, the standard form of $1.0 * 10^{-10}$ is $0.0000000001$

3 0

2 years ago

What are the factors that influence the power input to the compressor?

Lena [83]

Answer:

option e is correct answer

5 0

2 years ago

A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl

lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively

M<em>c</em> = 1.2 kg/s and M<em>h = </em>2 kg/s

C<em>pc</em> = 4.18 kj/kg °c and C<em>ph</em> = 4.31 kj/kg °c

<u>Using effectiveness-NUT method</u>

<em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>

C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c

C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= C<em>min</em>/C<em>max</em>

C = 5.016/8.62 = 0.582

.<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>

Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW

.<em>3 Third, we determine the actual heat transfer rate, Q</em>

Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW

.<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε

ε<em> </em>= Q/Qmax

ε <em>= </em>303.66 kW/702.24 kW

ε = 0.432

.<em>5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>

NTU = $\\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )$

NTU = $\frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582 -1} )$

NTU = 0.661

<em>.6 sixth, we determine Heat Exchanger surface area, As</em>

From the question, the overall heat transfer coefficient U = 640 W/m²

As = $\frac{NTU C{min} }{U}$

As = $\frac{0.661 x 5016 W. °c }{640 W/m²}$

As = 5.18 m²

<em>.7 Finally, we determine the length of the heat exchanger, L</em>

L = $\frac{As}{\pi D}$

L = $\frac{5.18 m² }{\pi (0.015 m)}$

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0

3 years ago