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4 years ago

So far in your lifetime, about how much garbage have you contributed

Engineering

1 answer:

OverLord2011 [107]4 years ago

7 0

Answer:

About 6132kg

Explanation:

In average, a person makes 638kg of garbage per year, and I’m 14 y’ old

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The condensed Q formula may be used for operations in which the friction loss can be determined for:

yan [13]

The condensed Q formula may be used for operations in which the friction loss can be determined for a: 3, 4, or 5 inch hose.

<h3>What is a firehose friction loss?</h3>

A firehose friction loss can be defined as a measure of the effect of the resistance of water against the inner side of a firehose, which typically results in a pressure drop at the terminal end.

Generally, some of the factors that affect the resistance or friction in a firehose include:

Length of hose.

Age of hose.

Water flow (gpm)

Water turbulence

Gravity

Mathematically, the firehose friction loss can be calculated by using this formula:

FL = C × (Q/100)² × L/100.

Read more on friction loss here: brainly.com/question/17305262

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2 years ago

The output S/N at thereceiver must be greater than 40 dB. The audio signal has zero mean, maximum amplitude of 1, power of ½ Wan

abruzzese [7]

Given that,

The output signal at the receiver must be greater than 40 dB.

Maximum amplitude = 1

Bandwidth = 15 kHz

The power spectral density of white noise is

$\dfrac{N}{2}=10^{-10}\ W/Hz$

Power loss in channel= 50 dB

Suppose, Using DSB modulation

We need to calculate the power required

Using formula of power

$P_{L}_{dB}=10\log(P_{L})$

Put the value into the formula

$50=10\log(P_{L})$

$P_{L}=10^{5}\ W$

For DSB modulation,

Figure of merit = 1

We need to calculate the input signal

Using formula of FOM

$FOM=\dfrac{\dfrac{S_{o}}{N_{o}}}{\dfrac{S_{i}}{N_{i}}}$

$1=\dfrac{\dfrac{S_{o}}{N_{o}}}{\dfrac{S_{i}}{N_{i}}}$

$\dfrac{S_{i}}{N_{i}W}=\dfrac{S_{o}}{N_{o}}$

Put the value into the formula

$\dfrac{S_{i}}{2\times10^{-10}\times15\times10^{3}}$

$\dfrac{S_{i}}{30\times10^{-7}}$

$S_{i}$

$S_{i}=30\times10^{-3}$

We need to calculate the transmit power

Using formula of power transmit

$S_{i}=\dfrac{P_{t}}{P_{L}}$

$P_{t}=S_{i}\times P_{L}$

Put the value into the formula

$P_{t}=30\times10^{-3}\times10^{5}$

$P_{t}=3\ kW$

We need to calculate the needed bandwidth

Using formula of bandwidth for DSB modulation

$bandwidth=2W$

Put the value into the formula

$bandwidth =2\times15$

$bandwidth = 30\ kHz$

Hence, The transmit power is 3 kW.

The needed bandwidth is 30 kHz.

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Answer:

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Explanation:

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