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e-lub [12.9K]
3 years ago
11

Wireless Internet networks, including many used in homes, often make use of high-frequency radio waves. High-frequency waves are

useful because they can carry a lot of information. However, high-frequency waves are less capable of passing through objects than are low-frequency waves. As a result, waves traveling from a person's wireless laptop computer, for example, could be interrupted by objects between the computer and the modem.
Due to this limitation of high frequency waves, which of the following statements best explains why digital waves are commonly used in high-frequency wireless networks instead of analog waves?
A.
Analog waves cannot reach high enough amplitudes to be used in wireless networks.
B.
Analog waves cannot reach large enough wavelengths to be used in wireless networks.
C.
It is easier to correct small disruptions in digital waves than in analog waves.
D.
Small disruptions in analog waves are more easily corrected than in digital waves.
Physics
2 answers:
IgorLugansk [536]3 years ago
7 0

Answer:

The CORRECT answer is c

lutik1710 [3]3 years ago
4 0
The answer is d , small disruptions
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3 years ago
Newer telephone circuits, built during the last decade, offer higher quality because they were built using analog transmission.
hammer [34]

Answer:

Well, newer telephone circuits built during the last decade are based on the digital transmission, not on the analog transmission. So it's the digital transmission circuit that has made the higher quality. Digital circuits converts the voice signals into the binary codes which is then translated again into the voice signal at the receiving end.

The answer is false.

Explanation:

6 0
3 years ago
A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a
Makovka662 [10]

Answer:

0.51 m

Explanation:

Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.

Kinetic energy, KE=½kx²

Where k is spring constant and x is the compression of spring

Potential energy, PE=mgh

Where g is acceleration due to gravity, h is height and m is mass

Equating KE=PE

mgh=½kx²

Making x the subject of formula

x=\sqrt {\frac {2mgh}{k}}

Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then

x=\sqrt \frac {2*1300*9.81*10}{1000000}=0.50503465227646m\\x\approx 0.51 m

5 0
3 years ago
There are two tuning forks struck at the same time. One tuning fork is tuned to a frequency note of 500 Hz. The other is tuned t
fomenos

Answer: 26 beats in 2 seconds

Explanation:

The number of beats per second = frequency of tuning fork 1 - frequency of tuning fork 2.

Given :

- frequency of tuning fork 1 = 500Hz

- frequency of tuning fork 2 = 487Hz

Thus,

Beats per second = 500 - 487

= 13 beats per second

Therefore in two(2) seconds, you will have 2 x 13 = 26 seconds.

6 0
3 years ago
An electron in a vacuum is first accelerated by a voltage of 81700 V and then enters a region in which there is a uniform magnet
Vilka [71]

Answer:

Magnetic force is equal to 1.37\times 10^{-11}N

Explanation:

We have given electron is accelerated with a potential difference of 81700 volt.

Magnetic field B = 0.508 T

Angle between magnetic field and velocity \Theta =90^{0}

Mass of electron m=9.11\times 10^{-31}kg

Charge on electron e=1.6\times 10^{-19}C

By energy conservation.

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

\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=1.6\times 10^{-19}\times 81700

v=169.4\times 10^6m/sec

Magnetic force on electron

F=qvBsin\Theta

F=1.6\times 10^{-19}\times 169.4\times 10^6\times 0.508\times sin90^{\circ}

=1.37\times 10^{-11}N

8 0
3 years ago
Read 2 more answers
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