A questiom for the big carb himself no cap

A network address of 172.16.0.0 /12 has been given. Which of the following accurately describes this network? (select one or mor

ludmilkaskok [199]

Answer:

B and E is correct.

Explanation:

Given that network address

172.16.0.0/12

This is class B network type.

The ending of this network will be 172.31.255.255

In IP version 4 there are four following type of classes

1)Class A (0-127)

2)Class B (128-191)

3)Class C (191-223)

4)Class D(224-239)

5)Class E (240-255)

Generally class A,B,C and D are used.

So our options B and E is correct.

3 0

3 years ago

what is the expected life 1 inch diameter bar machined from AISI 1020 CD Steel is subjected to alternating bending stress betwee

Alexeev081 [22]

Answer:

1.287 *10⁷ cycles.

Explanation:

See attached pictures.

3 0

3 years ago

A museum has three rooms, each with a motion sensor (m0, m1, and m2) that outputs 1 when motion is detected. At night, the only

nordsb [41]

Answer:

a) see attachment

b) A= m0m1+ m1m2+ m0m2

see attachment for K-map

c) see attachment

Explanation:

a) see attachment for truth table

b) see attachment for k-map

A= m0m1+ m1m2+ m0m2

c) see attachment for gate level circuit

4 0

3 years ago

g A 12,000 m3/day treatment plant has a rectangular sedimentation basin with dimensions 12 meters wide, 3 meters deep, and 25 me

Nesterboy [21]

Answer:

The settling velocity of the particles (Vs) is greater than the overflow rate (V₀), thus the particles will settle out and will be removed.

Explanation:

Given;

volumetric flow rate of the treatment, Q₀ = 12,000 m³/day

length of the rectangular tank, L = 25 m

width of the tank, W = 12 m

height of the tank, H = 3 m

settling velocity of the particles, $V_s$ = 6 x 10⁻³ m/s

The overflow rate of the sediments are calculated as follows;

$V_o = \frac{Q_o}{A_s}$

where;

As is the surface area of the tank, m²

Q₀ is the flow rate, m³/s

As = 2LW + 2LH + 2WH

As = (2 x 25 x 12) + (2 x 25 x 3) + (2 x 12 x 3)

As = 822 m²

$Q_0 (m^3/s)= \frac{12,000 \ m^3}{day} \times \frac{1 \ day}{24 \ hr} \times \frac{1 \ hour}{60 \ \min} \times \frac{1 \ \min}{60 \ s} = \frac{12,000}{24 \times 60 \times 60} (m^3/s)= \frac{12,000}{86,400} \ m^3/s\\\\Q_o = 0.139 \ m^3/s$

The overflow rate;

$V_o = \frac{Q_0}{A_s} = \frac{0.139}{822} = 1.69 \times 10^{-4} \ m/s$

The settling velocity of the particles (Vs) is greater than the overflow rate (V₀), thus the particles will settle out and will be removed.

7 0

3 years ago

A guy named Kirby Had touched only a gear on the projector. Yet there was a out of the blue rip. Kirby did not mind at all and s

solong [7]

The guy named Kirby didn't touch the power button on the projector. If he touched means the projector will work fine.

<u>Explanation</u>:

Kirby touched the gear on the projector. He should check the projector carefully before operating the projector. But he didn't check properly.
There was a blue rip in the projector. Kirby fails to notice the blue rip. Without minding that he has started the projector. So the tape and the projector is not working well.
He should fix the blue rip and he should press the start button in the projector. The mistake is he didn't do that. This is the thing that made the projector ripped.

7 0

3 years ago