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Natasha_Volkova [10]

3 years ago

12

When one of the data sources used for incident decision making is coming from individual or aggregated log files, the management

of those sources becomes critical. What are some of the key activities associated with managing logs?

1 answer:

Studentka2010 [4]3 years ago

3 0

Answer:

.

Explanation:

You might be interested in

40 = 6(X^2) / 4 (tan(180/6)) solve for X

const2013 [10]

Answer:

x = 3.92

Explanation:

The question is given as ;

40 = 6x² / 4 tan { 180/6}

40 = 6x² / 4 tan {30°}

40 = 6x² / 2.309

40 * 2.309 = 6x²

92.38 = 6 x²

92.38/6 = x²

15.40 = x²

√15.40 = x

x = 3.92

3 0

3 years ago

A 5-mm-thick stainless steel strip (k = 21 W/m•K, rho = 8000 kg/m3, and cp = 570 J/kg•K) is being heat treated as it moves throu

Drupady [299]

Answer:

The temperature of the strip as it exits the furnace is 819.15 °C

Explanation:

The characteristic length of the strip is given by;

$L_c = \frac{V}{A} = \frac{LA}{2A} = \frac{5*10^{-3}}{2} = 0.0025 \ m$

The Biot number is given as;

$B_i = \frac{h L_c}{k}\\\\B_i = \frac{80*0.0025}{21} \\\\B_i = 0.00952$

$B_i$ < 0.1, thus apply lumped system approximation to determine the constant time for the process;

$\tau = \frac{\rho C_p V}{hA_s} = \frac{\rho C_p L_c}{h}\\\\\tau = \frac{8000* 570* 0.0025}{80}\\\\\tau = 142.5 s$

The time for the heating process is given as;

$t = \frac{d}{V} \\\\t = \frac{3 \ m}{0.01 \ m/s} = 300 s$

Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;

$T(t) = T_{ \infty} + (T_i -T_{\infty})e^{-t/ \tau}\\\\T(t) = 930 + (20 -930)e^{-300/ 142.5}\\\\T(t) = 930 + (-110.85)\\\\T_{(t)} = 819.15 \ ^0 C$

Therefore, the temperature of the strip as it exits the furnace is 819.15 °C

5 0

3 years ago

garik1379 [7]

Answer:

The answer is below

Explanation:

1)

$R_{AB}=(500+500)||(500+500)\\\\R_{AB}=1000||1000\\\\R_{AB}=\frac{1000*1000}{1000+1000} \\\\R_{AB}=500\ \Omega$

2)

$R_{AB}=1000\|(1000+1000+1000)\\\\R_{AB}=1000||3000\\\\R_{AB}=\frac{1000*3000}{1000+3000} \\\\R_{AB}=750\ \Omega$

3)

Because of the short, the resistance is zero.

$R_{AB}=0$

4)

$R_{AB}=940\ \Omega$

5)

$R_{AB}=2200||2200||(2200+2200)\\\\R_{AB}=2200||2200||4400\\\\\frac{1}{R_{AB}}=\frac{1}{2200} +\frac{1}{2200} +\frac{1}{4400} \\\\\frac{1}{R_{AB}}=\frac{5}{4400}\\\\R_{AB}=880$

6)

$R_{AB}=(220+100)||470||330\\\\R_{AB}=320||470||330\\\\\frac{1}{R_{AB}}=\frac{1}{320} +\frac{1}{470} +\frac{1}{330} \\\\R_{AB}=120.7\ \Omega$

5 0

3 years ago

Select the correct answer.

andre [41]

Answer:

A. energy transformations

Explanation:

8 0

2 years ago

The settlement analysis for a proposed structure indicates that the underlying clay layer will settle 7.5 cm in 3 years, and tha

trapecia [35]

Answer:

A) For a double drainage system the time taken for total settlement will differ from that of a single drainage system, because in a double drainage case clay water escapes from both sides unlike in a single drainage case. hence water and air will be be expulsed faster for a double drainage making it attain total settlement faster.

B) for only single = 12 years

Explanation:

Given data :

ultimate clay settlement = 32 cm

settlement of clay in 3 years = 7.5 cm

Cv ( coefficient of consolidation ) = 1.5 x 10^-4 cm^2 /s

A) For a double drainage system the time taken for total settlement will differ from that of a single drainage system, because in a double drainage case clay water escapes from both sides unlike in a single drainage case. hence water and air will be be expulsed faster for a double drainage making it attain total settlement faster.

note : Total settlement is the same in both drainage system.

<u>B) Determine how long it will take for 7.5 cm of settlement to occur if there is only single </u>

applying the relation below

Tv = Cv t / H^2

where Tv = time factor , d = thickness of layer , H = drainage path

Given that Tv and Cv are constant in both cases

t ∝ H^2 hence

$\frac{t1}{t2} = \frac{H^21}{H^22}$ ------ ( 1 )

t1 = time for single drainage for d meters , t2 = time for double drainage for d meters

equation 1 can be rewritten as

t1/t2 = d^2 / (d/2)^2

∴ t1 = 4t2

given that t2 = 3 years ( value gotten from question )

t1 = 4 * 3 = 12 years

7 0

3 years ago