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4 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 .

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During an ultrasound, sound waves are sent by a transducer through muscle tissue at a speed of 1300 m/s. Some of the sound waves

alina1380 [7]

$2.3 \times 10^{-5} \text { seconds }$ is the time taken by the transducer to detect the reflected waves from the metal fragment after they were first emitted

Option C

Explanation:

Given data:

speed, v = 1300 m/s

distance, d = 3.0 cm = $3.00 \times 10^{-2} \mathrm{m}$

We need to calculate the time taken by the transducer to detect the reflected waves from the metal fragment after they were first emitted.

As we know, the velocity is the ratio of distance and the time travelled by an object. The equation form is given by,

$\text {velocity, } v=\frac{\text {distance }(d)}{\text {time}(t)}$

By applying the given values to the above equation, we get

$1300=\frac{3.00 \times 10^{-2}}{t}$

$t=\frac{3.00 \times 10^{-2}}{1300}=0.002307 \times 10^{-2}=2.3 \times 10^{-5} \text { seconds }$

7 0

4 years ago

5. A stream of air at 12 bar and 900 K is mixed with another stream of air at 2 bar and 400 K with 2 times the mass flowrate. If

Softa [21]

Answer:

The anserrs to the question are

(a) The temperature would be 566.67 K and

(b) The pressure of the resulting air stream is 14 bar

Explanation:

Here the two streams of gases meet ad form a single stream

The steady-flow energy equation can be implemented at the mixing point of the to streams as follows

mₐhₐ₁ + mₙ hₙ₁ + Q° + W° = mₐhₐ₂ + mₙhₙ₂

Where the flow is adiabatic, we have

Q = 0 and W = 0 hence

mₐhₐ₁ + mₙ hₙ₁ = mₐhₐ₂ + mₙhₙ₂ where h = cp×T we have

mₐcpₐ×Tₐ + mₙcpₙ×Tₙ = mₐcpₐ×T + mₐcpₙ×T

Therefore the final temperature T is given by

$T = \frac{m_ac_{pa}T_a + m_nc_{pn}T_n}{m_ac_{pa} + m_nc_{pn}}$ for the same kind of gas cpₐ =cpₙ therefore

$T = \frac{m_aT_a + m_nT_n}{m_a + m_n}$ where mₐ and mₙ are mass flow rate

Therefore we have where Tₐ = = 900 K and Tₙ = 400 K and mₙ = 2mₐ gives

$T = \frac{m_a*900 + 2*m_a*400}{m_a + 2*m_a}$ = $T = \frac{900 + 800}{1 + 2}$ = 1700/3 = 566.67 K

From Dalton's law, the total pressure of a mixture of gases is equal to the sum of the partial pressures of the components of the mixture

Therefore the total pressure of the combined stream = pₐ + pₙ = p

= 12 bar + 2 bar = 14 bar

Stream pressure = 14 bar

4 0

3 years ago

A student measures the diameter of a small cylindrical object and gets the following readings: 4.32, 4.35, 4.31, 4.36, 4.37, 4.3

Zinaida [17]

Answer:

a. $\bar{d}=4.34 cm$

b. $\sigma=0.023 cm$

c. $\rho=(0.0089\pm 0.00058) kg/cm^{3}$

Explanation:

a) The average of this values is the sum each number divided by the total number of values.

$\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}$

$x_{i}$ is values of each diameter
N is the total number of values. N=6

$\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}$

$\bar{d}=4.34 cm$

b) The standard deviation equations is:

$\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}$

If we put all this values in that equation we will get:

$\sigma=0.023 cm$

Then the mean diameter will be:

$\bar{d}=(4.34\pm 0.023)cm$

c) We know that the density is the mass divided by the volume (ρ = m/V)

and we know that the volume of a cylinder is: $V=\pi R^{2}h$

Then:

$\rho=\frac{m}{\pi R^{2}h}$

Using the values that we have, we can calculate the value of density:

$\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}$

We need to use propagation of error to find the error of the density.

$\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}$

δm is the error of the mass value.
δd is the error of the diameter value.
δh is the error of the length value.

Let's find each partial derivative:

1. $\frac{\partial\rho}{\partial m}=\frac{4m}{\pi d^{2}h}=\frac{4*1.66}{\pi 4.34^{2}*12.6}=0.0089$

2. $\frac{\partial\rho}{\partial d}=-\frac{8m}{\pi d^{3}h}=-\frac{8*1.66}{\pi 4.34^{3}*12.6}=-0.004$

3. $\frac{\partial\rho}{\partial h}=-\frac{4m}{\pi d^{2}h^{2}}=-\frac{4*1.66}{\pi 4.34^{2}*12.6^{2}}=-0.00071$

Therefore:

$\delta\rho=\sqrt{\left(0.0089)^{2}*0.05^{2}+\left(-0.004)^{2}*0.023^{2}+\left(-0.00071)^{2}*0.5^{2}}$

$\delta\rho=0.00058$

So the density is:

$\rho=(0.0089\pm 0.00058) kg/cm^{3}$

I hope it helps you!

3 0

3 years ago

A metal ball is just able to pass through a metal ring. When Anette increases the temperature of the ball, however, it will not

Natasha2012 [34]

Answer:

The size of the hole will increase

Explanation:

Heat is a form of energy that depends to a great extent on the temperature of the system,when a solid is heated there will be increases in it's length,area and volume otherwise known as linear,beta and superficial expansivities respectively.When Annette increased the temperature of the ring she supplied heat to the ring that led to the expansion of the ring,what really happened was that she increased the area expansivities of the ring which is the increase in area per unit area per degree rise in temperature,at such the interior diameter of the ring increase to such an extent that the ball passes through it.

5 0

3 years ago

A 6 cm long spring extends to 9 cm when a 1 kg load is suspended from it. What would be its length if a 2 kg load were suspended

Ganezh [65]

The answer is D that is the right question

8 0

3 years ago