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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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Suppose a car approaches a hill and has an initial speed of
kvv77 [185]

Answer:

a) 1.73*10^5 J

b) 3645 N

Explanation:

106 km/h = 106 * 1000/3600 = 29.4 m/s

If KE = PE, then

mgh = 1/2mv²

gh = 1/2v²

h = v²/2g

h = 29.4² / 2 * 9.81

h = 864.36 / 19.62

h = 44.06 m

Loss of energy = mgΔh

E = 780 * 9.81 * (44.06 - 21.5)

E = 7651.8 * 22.56

E = 172624.6 J

Thus, the amount if energy lost is 1.73*10^5 J

Work done = Force * distance

Force = work done / distance

Force = 172624.6 / (21.5/sin27°)

Force = 172624.6 / 47.36

Force = 3645 N

5 0
3 years ago
What is 6.9 × 107 written in standard form?
n200080 [17]
Answer: 6.9x 107 in standard form is 69,000,000

7 0
3 years ago
Read 2 more answers
A horizontal force of magnitude 30.2 N pushes a block of mass 3.50 kg across a floor where the coefficient of kinetic friction i
marshall27 [118]

Answer:

A) 89.39 J

B) 30.39J

C) 23.8 J

Explanation:

We are given;

F = 30.2N

m = 3.5 kg

μ_k = 0.646

d = 2.96m

ΔEth (Block) = 35.2J

A) Work done by the applied force on the block-floor system is given as;

W = F•d

Thus, W = 30.2 x 2.96 = 89.39 J

B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;

ΔEth = μ_k•mgd

Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J

Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.

Thus,

ΔEth = ΔEth (Block) + ΔEth (floor)

Thus,

ΔEth (floor) = ΔEth - ΔEth (Block)

ΔEth (floor) = 65.59J - 35.2J = 30.39J

C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;

W = K + ΔEth

Therefore;

K = W - ΔEth

K = 89.39 - 65.59 = 23.8J

3 0
3 years ago
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Brrunno [24]

Answer:

A current can be induced in a conducting loop if it is exposed to a changing magnetic field. ... In other words, if the applied magnetic field is increasing, the current in the wire will flow in such a way that the magnetic field that it generates around the wire will decrease the applied magnetic field.

Explanation:

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