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aleksklad [387]

3 years ago

5

A telephone line has a signal-to-noise ratio of 1000 and a bandwidth of 4 KHz. What is the maximum data rate supported by this l

ine?

1 answer:

ivolga24 [154]3 years ago

5 0

Answer:

The maximum data rate supported by this line is 39900 bps

Explanation:

The maximum data rate supported by this line can be obtained using the formula below

c = W*log2(S/N+1)

where;

c is the maximum data rate supported by the line

W is the bandwidth = 4kHz

S/N+1 is the signal to noise ratio = 1001

c = 4*log2(1001)

c = 39868.9 ≅ 39900 bps

Therefore, the maximum data rate supported by this line is 39900 bps

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You measure the velocity of a drag racer that accelerates with constant acceleration. You want to plot the data and determine th

Ivenika [448]

Answer:

a) Linear equation

Explanation:

Definition of acceleration

$a=\frac{dv}{dt}\\$

if a=constant and we integrate the last equation

$v(t)=v_{o}+a*t$

So the relation between the time and the velocity is linear. If we plot the velocity in function of time, the plot is a line, and the acceleration is the slope of this line.

8 0

2 years ago

(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i

murzikaleks [220]

Answer:

a) $t = -\frac{ln(2)}{k}$

b) See the proof below

$A(t) = A_o 2^{-\frac{t}{T}}$

c) $t = 3T \frac{ln(2)}{ln(2)}= 3T$

Explanation:

Part a

For this case we have the following differential equation:

$\frac{dA}{dt}= kA$

With the initial condition $A(0) = A_o$

We can rewrite the differential equation like this:

$\frac{dA}{A} =k dt$

And if we integrate both sides we got:

$ln |A|= kt + c_1$

Where $c_1$ is a constant. If we apply exponential for both sides we got:

$A = e^{kt} e^c = C e^{kt}$

Using the initial condition $A(0) = A_o$ we got:

$A_o = C$

So then our solution for the differential equation is given by:

$A(t) = A_o e^{kt}$

For the half life we know that we need to find the value of t for where we have $A(t) = \frac{1}{2} A_o$ if we use this condition we have:

$\frac{1}{2} A_o = A_o e^{kt}$

$\frac{1}{2} = e^{kt}$

Applying natural log we have this:

$ln (\frac{1}{2}) = kt$

And then the value of t would be:

$t = \frac{ln (1/2)}{k}$

And using the fact that $ln(1/2) = -ln(2)$ we have this:

$t = -\frac{ln(2)}{k}$

Part b

For this case we need to show that the solution on part a can be written as:

$A(t) = A_o 2^{-t/T}$

For this case we have the following model:

$A(t) = A_o e^{kt}$

If we replace the value of k obtained from part a we got:

$k = -\frac{ln(2)}{T}$

$A(t) = A_o e^{-\frac{ln(2)}{T} t}$

And we can rewrite this expression like this:

$A(t) = A_o e^{ln(2) (-\frac{t}{T})}$

And we can cancel the exponential with the natural log and we have this:

$A(t) = A_o 2^{-\frac{t}{T}}$

Part c

For this case we want to find the value of t when we have remaining $\frac{A_o}{8}$

So we can use the following equation:

$\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}$

Simplifying we got:

$\frac{1}{8} = 2^{-\frac{t}{T}}$

We can apply natural log on both sides and we got:

$ln(\frac{1}{8}) = -\frac{t}{T} ln(2)$

And if we solve for t we got:

$t = T \frac{ln(8)}{ln(2)}$

We can rewrite this expression like this:

$t = T \frac{ln(2^3)}{ln(2)}$

Using properties of natural logs we got:

$t = 3T \frac{ln(2)}{ln(2)}= 3T$

8 0

3 years ago

Please I need help with this :(

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2 years ago

You decide to clean the bathroom. You notice that the shower is covered in a strange green slime. You decide to try to get rid o

topjm [15]

The lemonade juice and water.

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3 years ago

Molecules have properties that are different than ____________that form them

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5 0

3 years ago

Read 2 more answers