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Greeley [361]
3 years ago
14

A network administrator is setting up a web server for a small advertising office and is concerned with data availability. The a

dministrator wishes to implement disk fault tolerance using the minimum number of disks required. Which RAID level should the administrator choose?
A) RAID 0B) RAID 1
C) RAID 5
D) RAID 6
Physics
1 answer:
lianna [129]3 years ago
5 0

RAID 1 level should be  choosed by the administrator.

B) RAID 1

<u>Explanation:</u>

RAID 1 is commonly utilized with a couple of plates, however, it should be possible with additional, and would indistinguishably reflect/duplicate the information similarly over all the drives in the exhibit. RAID 1  requires at least two physical drives, as information is composed at the same time to two spots.

The drives are basically identical representations of one another, so on the off chance that one drive comes up short, the other one can dominate and give access to the information that is put away on that drive. The purpose of RAID 1 is principally for repetition, as you can totally lose a drive, yet at the same time keep awake and running off the extra drive(s).

Likewise, at least two circles are required for RAID 1 equipment usage. With programming RAID 1, rather than two physical circles, information can be reflected between volumes on a solitary plate.

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tino4ka555 [31]

Answer:\mu=0.48

Explanation:

Given

inclination \theta =31^{\circ}

Acceleration of object=0.99 m/s^2

Now using FBD

mg\sin \theta -f_r=ma

mg\sin \theta -\mu mg\cos \theta =ma

a=g\sin \theta -\mu g\cos \theta

0.99=5.04-\mu 8.4

\mu 8.4=4.057

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3 years ago
A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected
gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected  = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta

Put the value into the fomrula

5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45

15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}

15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}....(I)

Along y -axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta

Put the value into the formula

0+0=5\times v_{1}\sin30-3\times v_{2}\sin45

\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0...(II)

From equation (I) and (II)

v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}

v_{1}=2.19\ m/s

Put the value of v₁ in equation (I)

\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0

v_{2}=\dfrac{5.475\times\sqrt{2}}{3}

v_{2}=2.58\ m/s

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0
3 years ago
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Margarita [4]

Answer:

false

Explanation:

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Answer: Skilled manpower is essential to carry out several development activities.

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Answer:

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