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allsm [11]
3 years ago
11

Given the following access-control list: access-list 90 deny 10.168.7.0 0.0.0.255 access-list 90 permit host 10.168.7.10 access-

list 90 permit host 10.168.7.11 access-list 90 permit host 10.168.7.12 access-list 90 deny any What would be the end result if the network security engineer applies access-list 90 to the inbound interface of FastEthernet 0/3? Group of answer choices A. The access-list will permit host 10.168.7.10 to pass through the router. B. The access-list will permit host 10.168.7.15 to pass through the router. C. Denial of service for this network because the rules are not placed in the correct order. D. Both hosts 10.168.7.10 and 10.168.7.11 will be permitted. E. The access-list will generate an error message.
Engineering
1 answer:
Yuri [45]3 years ago
7 0

Answer:

D. Both hosts 10.168.7.10 and 10.168.7.11 will be permitted

Explanation:

access-list 90

deny 10.168.7.0 0.0.0.255

permit 10.168.7.10

permit 10.168.7.11

permit 10.168.7.12

deny any

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Consider a 1.80-m-tall man standing vertically in water and completely submerged in a pool. Determine the difference between the
defon

Answer:

17.658 kPa

Explanation:

The hydrostatic pressure of a fluid is the weight of a column of that fluid divided by the base of that column.

P = \frac{weight}{base}

Also, the weight of a column is its volume multiplied by it's density and the acceleration of gravity:

weight = \delta * v * g

Meanwhile, the volume of a column is the area of the base multiplied by the height:

V = base * h

Replacing:

P = \frac{\delta * base * h * g}{base}

The base cancels out, so:

P = \delta * h * g

The pressure depends only on the height of the fluid column, the density of the fluid and the gravity.

If you have two point at different heights (or depths in the case of objects submerged in water) each point will have its own column of fluid exerting pressure on it. Since the density of the fluid and the acceleration of gravity are the same for both points (in the case of hydrostatics density is about constant for all points, it is not the case in the atmosphere), we can write:

\Delta P = \rho * g * (h1 - h2)

We do not know at what depth the man of this problem is, but it doesn't matter, because we know the difference in height of the two points of interes (h1 - h2) = 1.8 m. So:

\Delta P = 9.81 \frac{m}{s^{2} } * 1000 \frac{kg}{m^3} * 1.8 m = 17658 Pa = 17.658 kPa

4 0
3 years ago
A weight-lifting athlete raises a mass of 160 kg through a vertical distance of 1.4 m. What force did
Over [174]

Answer:

1568N

2195.2J

Explanation:

Given parameters:

Mass of the weight = 160kg

Distance  = 1.4m

Unknown:

Force applied to lift the weight = ?

Energy expended  = ?

Solution:

The force applied in moving a body with a given mass through a distance is the weight;

     Force applied  = mg

Where m is the mass

           g is the acceleration due to gravity

i.  Applied force = 160 x 9.8  = 1568N

ii. The energy used to lift the weight is given as;

     Energy  = mgh

h is the vertical distance

     Energy  = 1568 x 1.4  = 2195.2J

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3 years ago
Which of the following identifies the three main steps in the design process of an engineer?
ikadub [295]

Answer:

define the problem, do background research and specify requirements

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3 years ago
A cylinder 8 in. in diameter and 3 ft long is concentric with a pipe of 8.25 in internal diameter. Between the cylinder and the
yarga [219]

Answer:

What force is requiere to move the cylinder along

Explanation:

The kinematic viscosity of the ol is 0.006 fr2

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3 years ago
A 1-ft rod with a diameter of 0.5 in. is subjected to a tensile force of 1,300 lb and has an elongation of 0.009 in. The modulus
iragen [17]

Answer:

E = 8.83 kips

Explanation:

First, we determine the stress on the rod:

\sigma = \frac{F}{A}\\\\

where,

σ = stress = ?

F = Force Applied = 1300 lb

A = Cross-sectional Area of rod = 0.5\pi \frac{d^2}{4} = \pi \frac{(0.5\ in)^2}{4} = 0.1963\ in^2

Therefore,

\sigma = \frac{1300\ lb}{0.1963\ in^2} \\\\\sigma = 6.62\ kips

Now, we determine the strain:

strain = \epsilon = \frac{elongation}{original\ length} \\\\\epsilon = \frac{0.009\ in}{12\ in}\\\\\epsilon =  7.5\ x\ 10^{-4}

Now, the modulus of elasticity (E) is given as:

E = \frac{\sigma}{\epsilon}\\\\E = \frac{6.62\ kips}{7.5\ x\ 10^{-4}}

<u>E = 8.83 kips</u>

7 0
3 years ago
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