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Karo-lina-s [1.5K]
4 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
Engineering
1 answer:
Finger [1]4 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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Chemical milling is used in an aircraft plant to create pockets in wing sections made of an aluminum alloy. The starting thickne
Lelu [443]

Answer:

a) metal removal rate is 1915.37 mm³/min

b) the time required to etch to the specified depth is 500 min or 8.333 hrs

Explanation:

Given the data in the question;

starting thickness of one work part of interest = 20 mm

depth of series of rectangular-shaped pockets = 12 mm

dimension of pocket = 200 mm by 400 mm

radius of corners of each rectangle = 15 mm

penetration rate = 0.024 mm/minute

etch factor = 1.75

a)

To get the metal removal rate MRR;

The initial area will be smaller compare to the given dimensions of 200mm by 400mm and the metal removal rate would increase during the cut as area is increased. so'

A = 200 × 400 - ( 30 × 30 - ( π × 15² ) )

= 80000 - ( 900 - 707 )      

= 80000 - 193

A = 79807 mm²

Hence, metal removal rate MRR = penetration rate × A

MRR = 0.024 mm/minute × 79807 mm²

MRR = 1915.37 mm³/min

Therefore, metal removal rate is 1915.37 mm³/min

b) To get the time required to etch to the specified depth;

Time to machine ( etch ) =  depth of series of rectangular-shaped pockets / penetration rate

we substitute

Time to machine ( etch ) = 12 mm / 0.024 mm/minute

Time to machine ( etch ) = 500 min or 8.333 hrs

Therefore, the time required to etch to the specified depth is 500 min or 8.333 hrs

3 0
3 years ago
A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of
larisa86 [58]

Answer:

The field strength needed is 0.625 T

Explanation:

Given;

angular frequency, ω = 400 rpm = (2π /60) x (400) = 41.893 rad/s

area of the rectangular coil, A =  L x B = 0.0611 x 0.05 = 0.003055 m²

number of tuns of the coil, N = 300 turns

peak emf = 24 V

The peak emf is given by;

emf₀ = NABω

B = (emf₀ ) / (NA ω)

B = (24) / (300 x 0.003055 x 41.893)

B = 0.625 T

Therefore, the field strength needed is 0.625 T

4 0
3 years ago
Determine the reactor volume (assume a CSTR activated sludge aerobic reactor at steady state) required to treat 5 MGD of domesti
12345 [234]

Answer:

1.0MG

Explanation:

to solve this problem we use this formula

S₀-S/t = ksx --- (1)

the values have been given as

concentration = S₀ = 250mg

effluent concentration = S= 10mg

value of K = 0.04L/day

x = 3000 mg

when we put these values into this equation,

250-10/t = 0.04x10x3000

240/t = 1200

we cross multiply from this stage

240 = 1200t

t = 240/1200

t = 0.2

remember the question says that 5MGD is required to be treated

so the volume would be

v = 0.2x5

= 1.0 MG

4 0
3 years ago
Carbon resistors often come as a brown cylinder with colored bands. These colored bands can be read to determine the manufacture
alexandr1967 [171]

Answer:

a) 4.7 kΩ, +/- 5%

b) 2.0 MΩ, +/- 20%

Explanation:

a) If the resistor has the following combination of color bands:

1) Yellow = 1st digit = 4

2) Violet = 2nd digit = 7

3) Red = multiplier = 10e2

4) Gold = tolerance = +/- 5%

this means that the resistor has 4700 Ω (or 4.7 kΩ), with 5% tolerance.

b) Repeating the process for the following combination of color bands:

1)  Red = 1st digit = 2

2) Black = 2nd digit = 0

3) Green = multiplier = 10e5

4) Nothing = tolerance = +/- 20%

This combination represents to a resistor of 2*10⁶ Ω (or 2.0 MΩ), with +/- 20% tolerance.

7 0
4 years ago
weight of 1000 pounds is suspended from two cables. The allowable stress in the cables is 1500 psi. Find the minimum diameter fo
kari74 [83]

Answer:

The minimum diameter for each cable should be 0.65 inches.

Explanation:

Since, the load is supported by two ropes and the allowable stress in each rope is 1500 psi. Therefore,

(1/2)(Weight/Cross Sectional Area) = Allowable Stress

Here,

Weight = 1000 lb

Cross-sectional area = πr²

where, r = minimum radius for each cable

(1/2)(1000 lb/πr²) = 1500 psi

500 lb/1500π psi = r²

r = √1.061 in²

r = 0.325 in

Now, for diameter:

Diameter = 2(radius) = 2r

Diameter = 2(0.325 in)

<u>Diameter = 0.65 in</u>

7 0
3 years ago
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