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3 years ago

6.03 Discussion: Then & Now - Safety

Engineering

1 answer:

34kurt3 years ago

6 0

Answer:

Information technology is important in our lives because it helps to deal with every day's dynamic things. Technology offers various tools to boost development and to exchange information. Both these things are the objective of IT to make tasks easier and to solve many problems.

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For a RC Low Pass filter, what is the approximate amplitude (as a decimal proportion of input amplitude) at 1/4 the cutoff frequ

Nimfa-mama [501]

Answer:

Explanation:

RC low Pass Filter is an electronic circuit that comprises of a resistor and capacitor and it functions to permit low-frequency signals depending on the design and reject the high-frequency signals above a given frequency known as the cutoff frequency.

From the diagram attached below:

$V_{in$ = the input signal

$V_o$ = the output signal

Since; $V_o$ is used across the capacitor C,

By using the potential divider equation we have:

$V_o =V_{in} \times \dfrac{X_C}{\sqrt{ R^2+X_C^2} }$

From above; $X_C$ = capacitive reactance ;

and The total impedance Z is illustrated as $Z = \sqrt{R^2+X_C^2}$

Thus;

$V_o =V_{in} \times \dfrac{X_C}{Z}$

Recall that;

$X_C = \dfrac{1}{2 \pi fC}$

Here; f denotes the frequency of the input signal

Since the cutoff frequency is related to the frequency at which the capacitive reactance and resistance are said to be the same, then:

The Cutoff frequency can be expressed as:

$F_C = \dfrac{1}{2 \pi RC}$

Also;

the frequency of input signal $f = \dfrac{F_c}{4}$

$f = \dfrac{1}{8 \pi RC}$

Hence;

$X_C = \dfrac{1}{2 \pi fC}$

$X_C = \dfrac{1}{2 \pi \times \dfrac{1}{8 \pi RC} \times C}$

$X_C = 4R$

Finally;

From $Z = \sqrt{R^2+X_C^2}$

$Z = \sqrt{R^2+(4R)^2}$

$Z = \sqrt{17R^2}$

Z $\simeq$ 4.12R

As such, the output will be:

$V_o =V_{in} \times \dfrac{X_C}{\sqrt{ R^2+X_C^2} }$

$V_o =V_{in} \times \dfrac{4R}{4.12R^2} }$

$V_o =0.97V_{in}$

So, if we regard $A_{in}$ to be the input amplitude, then $A_{out}$ i.e the output amplitude will also be $A_{out}$ = $0.97 A_{in}$

4 0

3 years ago

The minimum fresh air requirement of a residential building is specified to be 0.35 air changes per hour (ASHRAE, Standard 62, 1

Natalka [10]

We know that

$A=200m^2\\h=2.7m\\\upsilon= 5.5m/s\\\%_{air} = 35%$

So, the volume of the entire building is

$V=2.7*200 = 540m^3$

The flow capacity of the fan

$\dot{V} = \frac{0.35*540}{60}$

$\dot{V} = 3.15m^3/min$

As $1L=10^{-3}m^3,$

$\dot{V}=3150L/min$

For the other part we know

$\dot{V}=\frac{\pi d^2}{4}V$

The diameter is,

$d=\sqrt{\frac{4\dot{V}}{\pi V}}$

$d=\sqrt{\frac{4*3.15}{\pi* 5.5* 60}}$

<em>**Note 60 is for the minutes</em>

$d= 0.1101m$

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4 0

3 years ago

You just graduate from college and find a job (yah! Congrats!). You buy a new house and would like to install solar panels for t

Lena [83]

Answer: Number of panels needed is 27

Daily energy per 250 W panel is 1.1kWhr

Daily house energy used is 30kWhr

Explanation: To get 27 panel it is 250*30/1.1 =6818

6818/250= 27 panels needed daily

6 0

3 years ago

Design brief of circuit diagram

Lemur [1.5K]

A design brief should highlight the problem and specifications for the solution.

<h3>What is an electrical circuit?</h3>

An electrical circuit can be defined as an interconnection of different electrical components, in order to create a pathway for the flow of electric current (electrons) due to a driving voltage.

<h3>The components of an electrical circuit.</h3>

Generally, an electrical circuit comprises the following electrical components:

Resistors

Capacitors

Batteries

Transistors

Switches

Wires

<h3>What is a design brief?</h3>

A design brief can be defined as a short statement which is written to describe the need of an individual or business, and what type of solution will meet that need (problem) through the design of an electrical circuit.

This ultimately implies that, a design brief should highlight the problem and specifications for the solution.

Read more on design brief here: brainly.com/question/17848745

#SPJ1

5 0

2 years ago

Car A starts from rest at t = 0 and travels along a straight road with a constant acceleration of 6 ft/s2 until it reaches a spe

taurus [48]

The distance traveled by car A when they pass each other is 1071m

<h3>Calculations and Parameters</h3>

Given:

t=0

constant acceleration= 6 ft/s^2

speed= 80 ft/s^2

$S_A= \frac{at1^2}{2} + V_At\\S_B= V_B (t1 +t)$

= $2 * (\frac{27}{2})^{2}/2 + 27 * 32.9$

= 1071m