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NARA [144]
3 years ago
13

An engineer wants to set up simple password protection with no usernames for some switches in a lab, for the purpose of keeping

curious co-workers from logging into the lab switches from their desktop PCs. Which of the following commands would be a useful part of that configuration?
A. A login vty mode subcommand
B. A password password console subcommand
C. A login local vty subcommand
D. A transport input ssh vty subcommand
Physics
1 answer:
Andre45 [30]3 years ago
8 0

Answer:

A. A login vty mode subcommand

Explanation:

since we are protecting co-workers from connecting to the switches from their desktop PCs, we would need a Telnet line which is used to connect to devices remotely from other network devices on the same network segment as the device we want to connect to. A login local vty subcommand configures a local username for login access but since our design constraint is to configure without usernames, option A is the correct answer.

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How does Newtons first law of motion relate to planetary motion
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3 0
3 years ago
A cyclist is riding a bicycle at a speed of 22 mph on a horizontal road. The distance between the axles is 42 in., and the mass
stealth61 [152]

Answer:

The shortest distance is  S = 24.86 ft

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

   The speed of the bicycle is v_b = 22\ mph = 22 * \frac{5280}{3600}   =  32.26 ft/s

     The distance between the axial is  d = 42 \ in

The mass center of the cyclist and the bicycle is m_c = 26 \ in  behind the front axle

The mass center of the cyclist and the bicycle is m_h = 40 \ in above the ground

   For the bicycle not to be thrown over the

     Momentum about the back wheel must be zero so

                \sum _B = 0

=>             mg (26) = ma(40)

=>             a = \frac{26}{40} g

Here  g = 32.2 ft/s^2

     So     a =  \frac{26}{40} (32.2)

             a =  20.93 ft/s^2

Apply the equation of motion to this motion we have

       v^2 = u^2 + 2as

 Where  u = 32.26 ft /s

             and v = 0 since the bicycle is coming to a stop

        v^2 = (32.26)^2 - 2(20.93) S

=>      S = 24.86 ft        

                 

5 0
3 years ago
A particle with charge 8 µC is located on the x-axis at the point −10 cm , and a second particle with charge 3 µC is placed on t
bixtya [17]

Answer:Force on -7 uC charge due to charge placed at x = - 10cm

now we will have

towards left

similarly force due to -5 uC charge placed at x = 6 cm

now we will have

towards left

Now net force on 7 uC charge is given as

towards left

Explanation:

6 0
3 years ago
Why would you have trouble breathing at high altitudes?
monitta

Answer:

A. It is colder at the top of a mountain

Explanation:

4 0
3 years ago
ball A of mass 5.0 kg moving at 20 meters per second collides with ball B of unknown mass moving at 10. meters per second in the
antiseptic1488 [7]

Answer:The mass of ball B is 10 kg.

Explanation;

Mass of ball A = M_A=5 kg

Velocity of the ball A before collision:U_A=20 m/s

Velocity of ball A after collision=V_A=10 m/s

Mass of ball B= M_B

Velocity of the ball B before collision:U_B=10 m/s

Velocity of ball B after collision=V_B=15 m/s

M_AV_A+M_BV_B=M_AU_A+M_BU_B

5 kg\times 10 m/s+M_B\times 15=5 kg\times 20m/s+M_B\times 10m/s

M_B=10kg

The mass of ball B is 10 kg.

8 0
3 years ago
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