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Karo-lina-s [1.5K]

3 years ago

8

When you are configuring data deduplication, you must choose a usage type for the volume you are configuring. Which of the follo

wing are valid data deduplication usage types? (Choose all that apply.)General purpose file serverDatabase serverVirtual Desktop InfrastructureVirtualized Backup Server

1 answer:

Finger [1]3 years ago

6 0

Answer:

1. General purpose file server.

2. Virtual Desktop Infrastructure Server.

3. Virtualized Backup Server.

All except : Database server

Explanation:

As a simple definition, we can tell, data deduplication is an elimination of redundant data in data set and storing only one copy of the same data. It is done by identifying double byte patterns through data analysis, removing double data and replacing it with reference pointed to stored, single piece of data.

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Ma poate ajuta cineva?

kari74 [83]

Da, sigur. cu ce ai nevoie de ajutor?

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3 years ago

Military glorification through mosaics, relief carvings and triumphant arches are often associated with?

emmasim [6.3K]

Military glorification through mosaics, relief carvings and triumphant arches are often associated with roman architectural monument and is the correct choice.

<h3>What is a Monument?</h3>

This can be defined as a structure which is usually large and is used to commemorate the history of an influential person. This helps to serve as an example and also a reminder as to why the performance of good deeds are important

The use of military glorification through mosaics, relief carvings and triumphant arches were also often used by the Romans in other to commemorate war victories and the succession of a new ruler which is known as the emperor.

Read more about Roman monuments here brainly.com/question/15786258

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8 0

2 years ago

The following program includes fictional sets of the top 10 male and female baby names for the current year. Write a program tha

otez555 [7]

Answer:

Define Variables and Use List methods to do the following

Explanation:

#<em>Conjoins two lists together</em>

all_names = male_names.union(female_names)

#<em>Finds the names that appear in both lists, just returns those</em>

neutral_names = male_names.intersection(female_names)

#<em>Returns names that are NOT in both lists</em>

specific_names = male_names.symmetric_difference(female_names)

3 0

3 years ago

Read 2 more answers

A counter-flow double pipe heat exchanger is heat heat water from 20 degrees Celsius to 80 degrees Celsius at the rate of 1.2 kg

lakkis [162]

Answer:

$L=107.6m$

Explanation:

Cold water in: $m_{c}=1.2kg/s, C_{c}=4.18kJ/kg\°C, T_{c,in}=20\°C, T_{c,out}=80\°C$

Hot water in: $m_{h}=2kg/s, C_{h}=4.18kJ/kg\°C, T_{h,in}=160\°C, T_{h,out}=?\°C$

$D=1.5cm=0.015m, U=649W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2},L=?m$

Step 1: Determine the rate of heat transfer in the heat exchanger

$Q=m_{c}C_{c}(T_{c,out}-T_{c,in})$

$Q=1.2*4.18*(80-20)$

$Q=1.2*4.18*(80-20)$

$Q=300.96kW$

Step 2: Determine outlet temperature of hot water

$Q=m_{h}C_{h}(T_{h,in}-T_{h,out})$

$300.96=2*4.18*(160-T_{h,out})$

$T_{h,out}=124\°C$

Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)

$dT_{1}=T_{h,in}-T_{c,out}$

$dT_{1}=160-80$

$dT_{1}=80\°C$

$dT_{2}=T_{h,out}-T_{c,in}$

$dT_{2}=124-20$

$dT_{2}=104\°C$

$LMTD = \frac{dT_{2}-dT_{1}}{ln(\frac{dT_{2}}{dT_{1}})}$

$LMTD = \frac{104-80}{ln(\frac{104}{80})}$

$LMTD = \frac{24}{ln(1.3)}$

$LMTD = 91.48\°C$

Step 4: Determine required surface area of heat exchanger

$Q=UA_{s}LMTD$

$300.96*10^{3}=649*A_{s}*91.48$

$A_{s}=5.07m^{2}$

Step 5: Determine length of heat exchanger

$A_{s}=piDL$

$5.07=pi*0.015*L$

$L=107.57m$

7 0

3 years ago

t is desired to produce aligned carbon fiber-epoxy matric composite having a longitudinal tensile strength of 800 MPa. Given (a)

Nesterboy [21]

Answer:

The value of critical length = 3.46 mm

The value of volume of fraction of fibers = 0.43

Explanation:

Given data

$\sigma_{T}$ = 800 M pa

D = 0.017 mm

L = 2.3 mm

$\sigma_{f}$ = 5500 M pa

$\sigma_{m}$ = 18 M pa

$\sigma_c$ = 13.5 M pa

(a) Critical fiber length is given by

$L_{c} = \sigma_{f} (\frac{D}{2 \sigma_{c} } )$

Put all the values in above equation we get

$L_{c} =5500 (\frac{0.017}{(2) (13.5)} )$

$L_{c} = 3.46$ mm

This is the value of critical length.

(b).Since this critical length is greater than fiber length Than the volume fraction of fibers is given by

$V_{f} = \frac{\sigma_T - \sigma_m}{\frac{L\sigma_c}{D} - \sigma_m }$

Put all the values in above formula we get

$V_{f} = \frac{800-18}{\frac{(2.3)(13.5)}{0.017} - 18 }$

$V_{f}$ = 0.43

This is the value of volume of fraction of fibers.

3 0

3 years ago