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NARA [144]
4 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]4 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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Prove that..<br>please help<br>​
GaryK [48]

\large{ \boxed{ \bf{ \color{red}{Universal \: law \: of \: gravitation}}}}

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The forces along the line joining the centre of the two objects.

❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.

According to universal law of gravitation,

\large{ \longrightarrow{ \rm{F \propto m_1 m_2}}}

Also,

\large{ \longrightarrow{ \rm{ F \propto  \dfrac{1}{ {d}^{2} } }}}

Combining both, We will get

\large{ \longrightarrow{ \rm{F  \propto  \dfrac{ m_1 m_2}{ {d}^{2}}}}}

Or, We can write it as,

\large{ \longrightarrow{ \rm{F  \propto  \:  G \dfrac{ m_1 m_2}{ {d}^{2} }}}}

Where, G is the constant of proportionality and it is called 'Universal Gravitational constant'.

☯️ Hence, derived !!

<u>━━━━━━━━━━━━━━━━━━━━</u>

8 0
3 years ago
A 4.0 cm object is 6.0 cm from a convex mirror that has a focal length of 8.0 cm. What is the distance of the image from the mir
ioda

Mirror formula

1/f = (1/p) + (1/q)    [ f= focal length  ,  p= object distance ,  q = image distance ]

as given convex mirror so

distance of object =p= 6cm

focal length =  -8cm

to find q=?

1/q =( 1/f) - (1/ p)

1/q = (1/ 8) -(1/6)

1/q =-1/24

q=-24  ( negative sign shows image is virtual and behind the mirror)

5 0
4 years ago
What is the validity of v=2Rg​
Anarel [89]

There's no possible way to think about the validity of that equation until we know what the variables v, R, and g represent.

8 0
3 years ago
On the modern periodic table, elements are ordered according to what?
hram777 [196]

Their atomic number, or number of protons.

3 0
3 years ago
Nick, a 60 kg physics student, wants to go bungee jumping, but doesn't have a bungee cord. He finds a 15 m long, strong spring (
vitfil [10]

Answer:

h = 24.81 m

Explanation:

Given:-

- The mass of the student, m = 60 kg

- The length of the spring, L = 15 m

- The spring constant, k = 60 N/m

Find:-

How far below the bridge is he hanging

Solution:-

- First realize that after the student attempts a bungee jump he oscillates violently ( dynamic motion ). After some time all the kinetic energy has been converted to Elastic and gravitational potential energy student is (stationary) and hanging down on one end of the spring.

- We will apply equilibrium condition on the student. We see that there are two forces acting on the student. The weight (W) of the student acting downward is in combat with the spring restoring force (Fs) acting upwards.

- Apply equilibrium condition in vertical direction:

                               Fs - W = 0

                               Fs = W

- The weight and spring force can be expressed as:

                               k*x = m*g

Where,     g : Gravitational acceleration constant = 9.81 m/s^2

                x : The extension of the spring from original position

- Solve for the extension (x) of the spring for this condition.

                              x = m*g / k

- Plug in the values and evaluate:

                              x = (60 kg)*(9.81 m/s^2) / (60 N/m)

                              x = 9.81 m

- The spring extends for about 9.81 m from its original length. So the distance (h) from edge of the bridge would be:

                              h = L + x

                              h = 15 + 9.81

                              h = 24.81 m

4 0
3 years ago
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