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blondinia [14]
3 years ago
14

Which option identifies the step represented in the following scenario?

Engineering
1 answer:
Zolol [24]3 years ago
5 0

Answer:

Solve the problem

Explanation:

The statement is : The team requests bids from a vendor to offer alternatives to raw materials that are too expensive for the proposed

price of the tablet.

The team wanted vendors to submit bids so that it can choose an alternative to raw materials to replace those that were expensive as compared to previous proposed price of a tablet. This is a process of solving the problem to identify a better solution.

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The percentage modulation of AM changes from 50% to 70%. Originally at 50% modulation, carrier power was 70 W. Now, determine th
adoni [48]

Answer:

What is percentage modulation in AM?

The percent modulation is defined as the ratio of the actual frequency deviation produced by the modulating signal to the maximum allowable frequency deviation.

3 0
3 years ago
Quadrilateral ABCD is a rectangle.<br> If m ZADB = 7k + 60 and mZCDB = -5k + 40, find mZCBD.
yuradex [85]

Hope this helps...........

8 0
3 years ago
The basic concept of feedback control is that an error must exist before some corrective action can be made?
weqwewe [10]

Answer:

The correct answer is True.

Explanation:

The feedback control system implies that to make a feedback, there must first be an error, otherwise there will be nothing to correct.

This system works so that there is an output that is controlled through a signal.

This signal will be feedback and it will signal an error which will be detected by a controller that will allow entry into the system.

In basic words, this system processes signals, samples them in the form of an output, and re-enters them feedback to detect the error signal.  

7 0
3 years ago
A train starts from rest at station A and accelerates at 0.4 m/s^2 for 60 s. Afterwards it travels with a constant velocity for
mash [69]
<h3><u>The distance between the two stations is</u><u> </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>

\\

Explanation:

<h2>Given:</h2>

a_1 \:=\:0.4\:m/s²

t_1 \:=\:60\:s

v_{i1} \:=\:0\:m/s

a_2 \:=\:0\:m/s²

t_2 \:=\:25\:min\:=\:1500\:s

a_3 \:=\:-0.8\:m/s²

v_{f3} \:=\:0\:m/s

\\

<h2>Required:</h2>

Distance from Station A to Station B

\\

<h2>Equation:</h2>

a\:=\:\frac{v_f\:-\:v_i}{t}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v\:=\:\frac{d}{t}

\\

<h2>Solution:</h2><h3>Distance when a = 0.4 m/s²</h3>

Solve for v_{f1}

a\:=\:\frac{v_f\:-\:v_i}{t}

0.4\:m/s²\:=\:\frac{v_f\:-\:0\:m/s}{60\:s}

24\:m/s\:=\:v_f\:-\:0\:m/s

v_f\:=\:24\:m/s

\\

Solve for v_{ave1}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v_{ave}\:=\:\frac{0\:m/s\:+\:24\:m/s}{2}

v_{ave}\:=\:12\:m/s

\\

Solve for d_1

v\:=\:\frac{d}{t}

12\:m/s\:=\:\frac{d}{60\:s}

720\:m\:=\:d

d_1\:=\:720\:m

\\

<h3>Distance when a = 0 m/s²</h3>

v_{f1}\:=\:v_{i2}

v_{i2}\:=\:24\:m/s

\\

Solve for v_{f2}

a\:=\:\frac{v_f\:-\:v_i}{t}

0\:m/s²\:=\:\frac{v_f\:-\:24\:m/s}{1500\:s}

0\:=\:v_f\:-\:24\:m/s

v_f\:=\:24\:m/s

\\

Solve for v_{ave2}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v_{ave}\:=\:\frac{24\:m/s\:+\:24\:m/s}{2}

v_{ave}\:=\:24\:m/s

\\

Solve for d_2

v\:=\:\frac{d}{t}

24\:m/s\:=\:\frac{d}{1500\:s}

36,000\:m\:=\:d

d_2\:=\:36,000\:m

\\

<h3>Distance when a = -0.8 m/s²</h3>

v_{f2}\:=\:v_{i3}

v_{i3}\:=\:24\:m/s

\\

Solve for v_{f3}

a\:=\:\frac{v_f\:-\:v_i}{t}

-0.8\:m/s²\:=\:\frac{0\:-\:24\:m/s}{t}

(t)(-0.8\:m/s²)\:=\:-24\:m/s

t\:=\:\frac{-24\:m/s}{-0.8\:m/s²}

t\:=\:30\:s

\\

Solve for v_{ave3}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v_{ave}\:=\:\frac{24\:m/s\:+\:0\:m/s}{2}

v_{ave}\:=\:12\:m/s

\\

Solve for d_3

v\:=\:\frac{d}{t}

12\:m/s\:=\:\frac{d}{30\:s}

360\:m\:=\:d

d_3\:=\:360\:m

\\

<h3>Total Distance from Station A to Station B</h3>

d\:= \:d_1\:+\:d_2\:+\:d_3

d\:= \:720\:m\:+\:36,000\:m\:+\:360\:m

d\:= \:37,080\:m

d\:= \:37.08\:km

\\

<h2>Final Answer:</h2><h3><u>The distance between the two stations is </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>
7 0
3 years ago
Express the Internal Energy and Entropy as a Function of T and V for a homogeneous fluid. Develop the same relations using the i
DedPeter [7]

Answer:

dU=C_{v} dT+(T(\frac{\beta }{\kappa })  -P)dV

dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV

Explanation:

The internal energy is equal to:

dU=C_{v} dT+(T(\frac{\delta P}{\delta T} )_{v} -P)dV

The entropy is equal to:

dS=C_{v} \frac{dT}{T} +(\frac{\delta P}{\delta T} )_{v} dV

If we write the pressure derivative in terms of isothermal compresibility and volume expansivity, we have

\frac{\delta P}{\delta T}=\frac{\beta }{\kappa }

Replacing:

dU=C_{v} dT+(T(\frac{\beta }{\kappa })  -P)dV

dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV

4 0
3 years ago
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