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3 years ago

How much data can a digital wave send. (The maximum data)

1 answer:

3 0

... A digital signal sent over a telephone connection is
limited to about 60,000 bits (7,500 Bytes) per second.

... Data sent over "Category 5" and "Category 6" Ethernet cables
usually runs at either 100 megabits or 1 gigabit per second.

... Data sent over wideband microwave radio signals
is currently capable of up to 1 gigabit (1000 megabits)
per second.

... Fiber optic systems typically operate at 10 gigabits or
40 gigabits per second, but much higher rates have been
accomplished.

In 2010, Nippon Telegraph and Telephone Corporation (of Japan)
transferred 69,100 gigabits per second (69.1 Terabits) over a single
fiber that was about 150 miles long.

In 2012, the same company transferred 1 million gigabits per second
(1 petabit) over a single fiber about 31 miles long.
__________________________________

1 megabit = 1 million bits

1 gigabit = 1 billion bits = 1,000 megabits

1 Terabit = 1,000 gigabits = 1 million megabits

1 petabit = 1,000 Terabits = 1 million gigabits = 1 billion megabits

= 1 billion million bits .

You might be interested in

Which of the following is a problem that some people blame on technology?

seropon [69]

The correct answer is B, widespread pollution. If you look closely, you can see that the other answers are not problems at all, but benefits! :)

7 0

3 years ago

Read 2 more answers

Classify the properties as extensive or intensive: mass density; color volume; total energy; temperature; melting point

m_a_m_a [10]

Answer:

Intensive properties

Density

Color

temperature

Melting point

Extensive properties

Mass

Volume

Total Energy

Explanation:

Intensive properties: In Physics, Intensive properties which are not depend of the amount of matter in a sample, It only depends of the type of matter, some examples of intensive properties are:

1. Density: It is a intensive property. It can explain better with a example: the water density is 1000 kg/m3, So if we have 1 liter or 1000 liters of water the density will be the same for the two samples.

2. Color: Solid sodium chloride is white. If you have 2 samples the first recipient with 2 kilograms of NaCl and the second with 10 kilograms of NaCl. The color of the substance does not depend on the amount of the substance.

As was mentioned before the same theory is applied to temperature and melting point concepts.

On the other hand,

Extensive properties are properties of the matter which depend on the amount of matter that is present in the system or sample. some examples are:

1. Mass: It is a property that measures the amount of matter that an object contains. For example, 10 kilograms of solid Copper contains a higher mass than 2 kilograms of the same metal.

2. Volume: It is a property which measures the space occupied by an object or a substance. For example, the space occupied by a glass of milk is lower than the space occupied by a bottle of milk, Then the volume of the glass of milk is lower than the volume of the bottle of milk.

3. Finally the total energy is contained in molecules and atoms that constituted systems so, if the amount of matter increases the number of molecules too, then the total energy will increase.

I hope it helps you.

6 0

3 years ago

A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red

Alik [6]

Answer:

Speed of physicist car is 0.036c or 1.08 x 10⁷ m/s .

Explanation:

Doppler Effect is defined as the change in frequency or wavelength of the wave as the source or/and observer moving away or towards each other.

In this case, the Doppler effect equation in terms of wavelength is :

$\lambda_{s} = \lambda_{o}\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }$ ......(1)

Here $\lambda_{s}$ is source wavelength, $\lambda_{o}$ is observed wavelength, v is speed of the observer and c is the speed of light.

Given :

Source wavelength, $\lambda_{s}$ = 660 nm = 660 x 10⁻⁹ m

Observed wavelength, $\lambda_{0}$ = 555 nm = 555 x 10⁻⁹ m

Substitute these values in the equation (1).

$555\times10^{-9} } = 660\times10^{-9} \sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }$

$\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = 0.84$

${\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = (0.84)^{2} = 0.7056$

$1-\frac{v}{c}=0.7056+0.7056\frac{v}{c}$

$\frac{v}{c}=\frac{0.2944}{8.056}$

$v = 0.036c=0.036\times3\times10^{8}$

v = 1.08 x 10⁷ m/s

8 0

3 years ago

oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student

puteri [66]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

3 0

3 years ago

A 40.0 kg wagon is towed up a hill inclined at 18.5 degrees with respect to the horizontal. The tow rope is parallel to the incl

sp2606 [1]

Answer:

7.9m/s

Explanation:

We are given that

Mass of wagon=40 kg

$\theta=18.5^{\circ}$

Tension= $140 N$

Initial velocity of wagon= $u=0$

Displacement=s=80 m

Net force applied on wagon= $F=T-mgsin\theta=140-40(9.8)sin18.5=15.7 N$

By using $g=9.8m/s^2$

$a=\frac{F}{a}=\frac{15.7}{40}=0.39m/s^2$

We know that

$v^2-u^2=2as$

Using the formula

$v^2=2\times 0.39\times 80$

$v=\sqrt{2\times 0.39\times 80}=7.9m/s$

5 0

3 years ago