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Allisa [31]
4 years ago
11

Typically a Switch operates at layer 2 of the OSI model. However, small organizations, such as a SOHO, can purchase a switch tha

t also interprets Layer 3 data and works much like a router. What is this device called
Physics
1 answer:
Romashka-Z-Leto [24]4 years ago
6 0

Answer:

Layer 3 switch.

Explanation:

A layer 3 switch carrys out both the function of a switch and a router. It acts as a switch that links all the devices that are on the same subnet or virtual LAN at lightning speeds and has IP routing intelligence built into it to carry out the function of a router. It can support routing protocols, check incoming packets, and can also carry out routing decisions based on the source and the destination addresses.

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The star nearest to our Sun is called Proxima Centauri. It has a surface temperature of approximately 3000 Kelvin and a luminosi
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Ummmm is there any options?

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3 years ago
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Suppose the voltage source for a series RL-circuit were given as V0sin(ωt) instead of V0cos(ωt). Write an expression for the cur
Gala2k [10]

Answer:

Explanation:

This is an RL circuit, therefore:

Impedance; z = \mathbf{\sqrt{R^2+L^2}}

\mathbf{z = \sqrt{R^2+(Lw)^2}}

Current amplitude

\mathbf{I_o = \dfrac{V_o}{z}} \\ \\  \mathbf{I_o = \dfrac{V_o}{\sqrt{R^2+L^2\omega ^2}}}

a)

Given that:

V_o = 1.9 \ V \\ \\ \omega= 51 \ rad/s\\\\ R = 21 \Omega \\ \\  L = 0.52 H

∴

I_o= \dfrac{1.9}{\sqrt{21^2+(0.52\times 51)^2}}

\mathbf{I_o= 0.0562}  \\ \\ \mathbf{I_o = 56.2 \ mA}

b)

Phase constant :

tan  \ \phi = \dfrac{L \omega}{R } \\ \\  tan  \ \phi = \dfrac{0.52 \times 51}{21} \\ \\  tan \phi = 1.263

\text{Phase constant : }\phi = tan^{-1} (1.263)   \\ \\  \phi = 51.6^0\\ \\\text{To radians} \phi  = 51.6 \times \dfrac{\pi}{180} \\ \\  \phi = 0.287 \pi \\ \\ \mathbf{\phi = 0.9 \ rad}

3 0
3 years ago
What happens to the ball's velocity while the ball is traveling upwards?
Bess [88]
If the ball does not have a propeller or jet engine on it, then it is an object
in free fall.  That means its downward speed grows by 9.8 m/s for every
second that it's in the air. 

If it happens to be traveling upward at the moment, then that won't last long. 
Its upward speed is decreasing by 9.8 m/s every second.  It will eventually
run out of upward gas and start moving downward.  At that instant, you might
say that the direction of its velocity has changed by 180 degrees.
7 0
4 years ago
Two spheres look identical and have the same mass. However, one is hollow and the other is solid. Describe in detail an experime
creativ13 [48]

Answer:

Put the two solid spheres on an inclined plane . Use a meter-stick to hold the spheres on the plane.  Release the two spheres at the same time and let down roll down.  Observe the two spheres as they roll down and repeat the steps.  The hollow sphere will roll last while the solid sphere will roll first. The hollow sphere has more rotational inertia than the solid sphere. This is because the mass of the hollow sphere is distributed farther from its center of gravity.

Explanation:

The description of the experiment for the two spheres is given below:

1. Put the two solid spheres on an inclined plane .

2. Use a meter-stick to hold the spheres on the plane.

3. Release the two spheres at the same time and let down roll down.

4. Observe the two spheres as they roll down and repeat the steps.

5. The hollow sphere will roll last while the solid sphere will roll first. The hollow sphere has more rotational inertia than the solid sphere. This is because the mass of the hollow sphere is distributed farther from its center of gravity.

7 0
3 years ago
A steel plate weighing 180 lb with center of gravity at point G is supported by a roller at point A, a bar DE, and a horizontal
melamori03 [73]

Answer:

Therefore, the force supported  by the hydraulic cylinder is = -70.01 lb

Thus, the force supported by the bar = -233.84 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Explanation:

The attached diagrams is shown in the file below:

From there; taking moments about point A

\sum M__A }=0

- 40 (180) - (80) F_E Cos 37° + 55  

-7200 + F_E (-80 Cos 37° + 55 sin 37° ) = 0

= - 7200 - 30.79  F_E

- 30.79  F_E = 7200

F_E = -\frac{7200}{30.79}

F_E  = -233.84 lb

Thus, the force supported by the bar = -233.84 lb

Taking the equilibrium of forces in the vertical direction

\sum f_y = 0

F_E sin 37 + F_A cos 20 - F_G = 0

- 233.84 sin 37 + F_A cos 20 - 180 = 0

F_A cos 20  = 320.73

F_A = \frac{320.73}{cos 20}

F_A = 341.31 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Taking the equilibrium of forces on the horizontal direction.

\sum f_y = 0

F_E cos 37 - F_{CB} + F_A sin 20° = 0

-233.84 cos 37 - F_{CB} + 341.31 sin 20° = 0  

-186.75 -  F_{CB} + 116.74 = 0

-  F_{CB} -70.01 = 0

-  F_{CB} = 70.01

F_{CB} = - 70.01 lb

Therefore, the force supported  by the hydraulic cylinder is = -70.01 lb

7 0
4 years ago
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