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2 years ago

if you're shopping for a rack switch, what component on the switch tells you it can be mounted to a rack?

1 answer:

6 0

if you're shopping for a rack switch, the component on the switch that tells you it can be mounted to a rack is the:

Rack ears

Rack ears are L-shaped objects that can be used to hold a rack switch firmly to the support walls. Rack ears are often found at the front panel of the rack which is to be mounted.

When a person purchases a rack switch and finds the rack ears there, it is a signal that the item can be secured firmly to rails. Sometimes, the rack ears also appear as extensions.

Learn more about rack ears here:

brainly.com/question/13318148

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Based on the situation above, choose the CORRECT type of error.

tatiyna

Answer:

what's your question I can't understand

5 0

3 years ago

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago

Lori’s family is on a road trip. They split their drive into the five legs listed in the table. Find the average velocity for ea

NARA [144]

It's not possible to answer the question exactly the way it's written.
That's because we don't know anything about the direction they
drive at any time during the trip.

You see, "velocity" is not just a word that you use for 'speed' when
you want to sound smart and technical, like this question is doing.
"Velocity" is a quantity that's made up of speed AND THE DIRECTION
of the motion. If you don't know the direction of the motion, then you
CAN'T tell the velocity, only the speed.

Here are the average speeds that Lori's family drove on each leg
of their trip:

Speed = (distance covered) / (time to cover the distance) .

Leg-A:
Speed = 15km/10min = 1.5 km/min

Leg-B:
Speed = 20km/15min = (1 and 1/3) km/min

Leg-C
Speed = 24km/12min = 2 km/min

Leg-D:
Speed = 36km/9min = 4 km/min

Leg-E:
Speed = 14km/14min = 1 km/min

From lowest speed to highest speed, they line up like this:

[Leg-E] ==> [Leg-B] ==> [Leg-A] ==> [Leg-C] ==> [Leg-D]
1.0 . . . . . . . . 1.3 . . . . . . . 1.5 . . . . . . . 2.0 . . . . . . . 4.0 . . . . km/minute

Whoever drove Leg-D should have been roundly chastised
and then abandoned by the rest of the family.  36 km in 9 minutes
(4 km per minute) is just about 149 miles per hour !

4 0

3 years ago

Read 2 more answers

When operated on a household 110.0 V line, typical hair dryers draw about 1450 W of power. The current can be modeled as a long,

dezoksy [38]

Answer:

0.075A

Explanation:

We can consider this system as a circuit, hence we can take the current from the formula for the electric power as follow

$P=IV\\I=\frac{P}{V}=\frac{110V}{1450W}=0.075A$