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horrorfan [7]
3 years ago
11

Which of the following is not a limitation of the current Internet? A) the continued reliance on cables and wires for connectivi

ty B) limited bandwidth, which causes congestion and cannot adequately handle video and voice traffic C) architectural restrictions, which stipulate that numerous requests for the same file must each be answered individually, slowing network performance D) the difficulty in expanding capacity by adding servers and clients
Physics
1 answer:
Alinara [238K]3 years ago
6 0

Answer:

D) the difficulty in expanding capacity by adding servers and clients

Explanation:

The limitation of the current Internet are as follows

1.The speed of the internet.When the number of user used the same internet then the speed of the internet decrease.

2.The band with is limited.

3.Continued reliance on cables and wires for connectivity

So the option d is correct.

D) the difficulty in expanding capacity by adding servers and clients

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If a charge at 60c flow in a conductor for 30 second then the current that flow in a conductor is​
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Explanation:

<h3>Given</h3>

- Charge = 60c

time = 30 sec

<h3>To find -</h3>

current

<h3>Solution </h3>

Current = Charge/time

I = V/T

I = 60/30

I = 2 ampere

More to know -

I = Current

V = Charge

T = Time

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3 years ago
Juan's mother drives 12.5 miles southwest to her favorite shopping mall. What is the velocity of her
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Calculate using the formula

Explanation:

velocity= displacement (m)/time(s)

1 mile =1.6km

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A passenger train travels 295 miles in the same amount of time it takes a freight train to travel 225 miles. The rate of the pas
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9.3

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3 years ago
The tungsten filament of a light bulb has an operating temperature of about 2 100 K. If the emitting area of the filament is 1.0
Murljashka [212]

Answer:

75 W

Explanation:

T = temperature of the filament = 2100 K

A = Emitting area of the filament = 1 cm² = 10⁻⁴ m²

e = Emissivity = 0.68

\sigma = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Using Stefan's law, Power output of the light bulb is given as

P = \sigma e AT^{4} \\P = (5.67\times10^{-8}) (0.68) (10^{-4}) (2100)^{4}\\P = 75 W

5 0
2 years ago
An electron emitted from a filament is travelling at 1.5 x 105 m/s when it enters an acceleration of an electron gun in a televi
Crank

Answer:

The acceleration of the electron is 1.457 x 10¹⁵ m/s².

Explanation:

Given;

initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s

distance traveled by the electron, d = 0.01 m

final velocity of the electron, v = 5.4 x 10⁶ m/s

The acceleration of the electron is calculated as;

v² = u² + 2ad

(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a

(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²

(2 x 0.01)a = 2.91375 x 10¹³

a = \frac{2.91375 \ \times \ 10^{13}}{2 \ \times \ 0.01} \\\\a = 1.457 \ \times \ 10^{15} \ m/s^2

Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².

7 0
3 years ago
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