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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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Answer:

a. 318.2k

b. 45.2kj

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Heat transfer rate to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.

See attachment for detailed analysis

7 0
3 years ago
A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic
DiKsa [7]

Answer:

The answer is below

Explanation:

1) The synchronous speed of an induction motor is the speed of the magnetic field of the stator. It is given by:

n_s=\frac{120f_s}{p}\\ Where\ p\ is \ the \ number\ of\ machine\ pole, f_s\ is\ the\ supply \ frequency\\and\ n_s\ is \ the \ synchronous\ speed(speed \ of\ stator\ magnetic \ field)\\Given: f_s=60\ Hz, p=4. Therefore\\\\n_s=\frac{120*60}{4}=1800\ rpm

2) The speed of the rotor is the motor speed. The slip is given by:

Slip=\frac{n_s-n_m}{n_s}. \\ n_m\ is\ the \ motor\ speed(rotor\ speed)\\Slip = 0.05, n_s= 1800\ rpm\\ \\0.05=\frac{1800-n_m}{1800}\\\\ 1800-n_m=90\\\\n_m=1800-90=1710\ rpm

3) The frequency of the rotor is given as:

f_r=slip*f_s\\f_r=0.04*60=2.4\ Hz

4) At standstill, the speed of the motor is 0, therefore the slip is 1.

The frequency of the rotor is given as:

f_r=slip*f_s\\f_r=1*60=60\ Hz

6 0
3 years ago
La extensión de un resumen debe estar en un rango de _________ del texto inicial. a)25 a 35 % b)10 a 20 % c)15 a 20 % d)20 a 25
mestny [16]

Answer:

b)

Explanation:

because it is correct

3 0
3 years ago
When two or more simple machines are combined they form
Volgvan
Compound machine is the answer
8 0
2 years ago
The water in a 25-m-deep reservoir is kept inside by a 140-m-wide wall whose cross section is an equilateral triangle as shown i
koban [17]

Answer:  (a) 9.00 Mega Newtons or 9.00 * 10^6 N

               (b)  17.1 m

Explanation:  The length of wall under the surface can be given by

                                            b=25m/sin(60)\\=28.867

The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.

F(resultant) = Pavg ( A) = (Patm +  \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N

Noting from the Bernoulli  equation that

Po/\rho g sin60 = 100000/1000 * 9.81* sin(60) = 11.77 m \\ \\

From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:

Yp = s+\frac{b}{2} +\frac{b^2}{s+\frac{b}{2}+Po/\rho g sin60}= 0+\frac{28.87}{2} +\frac{28.87^2}{0+\frac{28.87}{2}+100000 /1000 *9.81 sin60} = 17.1 m

Substituting the values gives us the the distance of the surface to be equal to = 17.1 m

7 0
3 years ago
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