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Vera_Pavlovna [14]
3 years ago
13

The most common example of a(n) ____ switched network is the conventional telephone system.

Physics
1 answer:
fenix001 [56]3 years ago
4 0
CIRCUIT Switched network has the conventional telephone system as the most common example.

Circuit switching<span> is a method of implementing a telecommunication network </span><span>in which two network nodes </span><span>establish a dedicated communications channel or circuit </span><span>through the network before the nodes may communicate. The circuit guarantees the full bandwidth of the channel and remains connected for the duration of the communication session by functioning as though the nodes were physically connected like an electrical circuit.</span>

In a conventional telephone system, when a call is made from one telephone to another, switches within the the telephone exchanges create a continuous wire circuit between the two units for as long as the call lasts.
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You cause a particle to move from point A, where the electric potential is 11.3 V, to point B, where the electric potential is −
BabaBlast [244]

Explanation:

The electric potential is the electric potential energy per unit of charge

V=\frac{U}{q}

Using this definition, we can calculate the electrostatic potential energy change between point A and B:

\Delta U=U_A-U_B\\\Delta U=qV_A-qV_B\\\Delta U=q(V_A-V_B)

Electron: q=-1.6*10^{-19}C

\Delta U=-1.6*10^{-19}C(11.3V+25.9V)\\\Delta U=-5.952*10^{-18}J

Proton: q=1.6*10^{-19}C

\Delta U=1.6*10^{-19}C(11.3V+25.9V)\\\Delta U=5.952*10^{-18}J

Neutral hydrogen atom: q=0

\Delta U=0

Singly ionized helium atom: q=1.6*10^{-19}C

\Delta U=1.6*10^{-19}C(11.3V+25.9V)\\\Delta U=5.952*10^{-18}J

8 0
3 years ago
Assuming that 70 percent of the Earth’s surface
Aneli [31]
We need to find the volume of a spherical shell with a radius of
6.37 million meters and a thickness of 0.95 mile.

The technically correct way to do this is to find the volume of the
outside of the shell, then find the volume of the inside of the shell,
and subtract the inside volume from the outside volume.  That's
the REAL way to do it.

But look.  This 'shell' (the 0.95 mile of water) is only about  1530 meters thick,
on a sphere with a radius of 6.37 million meters.  The depth of the water is like
0.024 percent of the radius !  There's not a whole lot of difference between the
sphere outside the water and the sphere inside it.

So I want to do this problem the easier way ... Let's say that the volume
of the water is going to be

                  (the surface area that it covers on the Earth)
         times
                  (the thickness of the coating of water) .

The area of a sphere is  4 pi Radius² .
That's
                         (4 pi) x (6.37 x 10⁶ m)²

                   =    (4 pi) x (40.58 x 10¹² m²)

We're only interested in 70% of the total surface area.

                   =   (0.7) x (4 pi) x (40.58 x 10¹²) m²

                   =            3.57 x 10¹⁴  square meters of Earth's surface.

The volume of the water covering that area is

               (the area) times (average depth of 0.95 mile) .

We have to change that 0.95 mile to meters.
The question reminds us that                         1 mile = 1609 meters .    
So the volume of the water is

                      (the area) times (0.95 x 1609 meters).

But we're not there yet.  The question isn't asking for the volume.
It's asking for the mass of the water. 
We're ready to get the volume in cubic meters.
We're supposed to know that each cubic meter is 1,000 liters,
   and the mass of 1 liter of water is 1 kilogram.
So each cubic meter of volume is 1,000 kilograms of mass.

Now we're ready to dump all the numbers into the machine and
turn the crank.  The mass of all this water will be

         (the surface area) x (0.95 x 1609 meters) x (1,000 kg/m³)   

  =    (3.57 x 10¹⁴  m²)  x   (1528.6 m)  x  (1,000 kg/m³)

  =            5.457 x 10²⁰ kilograms .

This is my answer, and I'm stickin to it.

But ... just like all the other problems you get in high school, the
answer doesn't matter.  The teacher doesn't need the answer,
and YOU don't need the answer.  The reason you got this problem
for an assignment is to give you practice in HOW TO FIND the
answer ... how to plan what you're going to do with the problem,
and then how to carry it out.

I don't know how much effort you put into this problem, but somewhere
along the way, you chickened out and posted it on Brainly.  So far, the
result of that decision was:  The person who got all the practice was ME.
I got the good stuff, and all YOU got was the answer.

I hope my work is clear enough that you can go through it, and pick up
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3 0
3 years ago
1) What would the average acceleration be for a car at a stoplight that speeds up to 20 m/s in 10 seconds (in m/s^2)
Alexxx [7]
1.)
Velocity is in m/s, and acceleration is in m/s^2 like you said. Because of this, we can calculate this by dividing the speed by the time it took to get to that speed.
(20 meters/second) / 10 seconds = 2 meters/ second^2

2.)
Same thing with the first one.
(100 meters/second) / 4 seconds = 25 meters / seconds^2
7 0
3 years ago
Read 2 more answers
A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h
hram777 [196]

Answer:

v_{2}=3.5 m/s

Explanation:

Using the conservation of energy we have:

\frac{1}{2}mv^{2}=mgh

Let's solve it for v:

v=\sqrt{2gh}

So the speed at the lowest point is v=7 m/s

Now, using the conservation of momentum we have:

m_{1}v_{1}=m_{2}v_{2}

v_{2}=\frac{1*7}{2}

Therefore the speed of the block after the collision is v_{2}=3.5 m/s

I hope it helps you!

       

8 0
3 years ago
what is the distance a train can travel if its speed is 20mph over a time of 5.6 hours (show all 3 steps)
erik [133]

Answer:

distance = 112 miles

Explanation:

its 12 miles every 0.6 in a hour

8 0
2 years ago
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