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 25 min. It then decelerates at 0.8 m/s^2 until it is brought to rest at station B.
Determine the distance between the stations. ?s^Delta s = _______.

Answers

Answer 1
The distance between the two stations is 37.08 km

[tex]\\[/tex]

Explanation:

Given:

[tex] a_1 \:=\:0.4\:m/s²[/tex]

[tex] t_1 \:=\:60\:s[/tex]

[tex] v_{i1} \:=\:0\:m/s[/tex]

[tex] a_2 \:=\:0\:m/s²[/tex]

[tex] t_2 \:=\:25\:min\:=\:1500\:s[/tex]

[tex] a_3 \:=\:-0.8\:m/s²[/tex]

[tex] v_{f3} \:=\:0\:m/s[/tex]

[tex]\\[/tex]

Required:

Distance from Station A to Station B

[tex]\\[/tex]

Equation:

[tex]a\:=\:\frac{v_f\:-\:v_i}{t}[/tex]

[tex]v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}[/tex]

[tex]v\:=\:\frac{d}{t}[/tex]

[tex]\\[/tex]

Solution:Distance when a = 0.4 m/s²

Solve for [tex]v_{f1}[/tex]

[tex]a\:=\:\frac{v_f\:-\:v_i}{t}[/tex]

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

[tex]24\:m/s\:=\:v_f\:-\:0\:m/s[/tex]

[tex]v_f\:=\:24\:m/s[/tex]

[tex]\\[/tex]

Solve for [tex]v_{ave1}[/tex]

[tex]v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}[/tex]

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

[tex]v_{ave}\:=\:12\:m/s[/tex]

[tex]\\[/tex]

Solve for [tex]d_1[/tex]

[tex]v\:=\:\frac{d}{t}[/tex]

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

[tex]720\:m\:=\:d[/tex]

[tex]d_1\:=\:720\:m[/tex]

[tex]\\[/tex]

Distance when a = 0 m/s²

[tex]v_{f1}\:=\:v_{i2}[/tex]

[tex]v_{i2}\:=\:24\:m/s[/tex]

[tex]\\[/tex]

Solve for [tex]v_{f2}[/tex]

[tex]a\:=\:\frac{v_f\:-\:v_i}{t}[/tex]

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

[tex]0\:=\:v_f\:-\:24\:m/s[/tex]

[tex]v_f\:=\:24\:m/s[/tex]

[tex]\\[/tex]

Solve for [tex]v_{ave2}[/tex]

[tex]v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}[/tex]

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

[tex]v_{ave}\:=\:24\:m/s[/tex]

[tex]\\[/tex]

Solve for [tex]d_2[/tex]

[tex]v\:=\:\frac{d}{t}[/tex]

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

[tex]36,000\:m\:=\:d[/tex]

[tex]d_2\:=\:36,000\:m[/tex]

[tex]\\[/tex]

Distance when a = -0.8 m/s²

[tex]v_{f2}\:=\:v_{i3}[/tex]

[tex]v_{i3}\:=\:24\:m/s[/tex]

[tex]\\[/tex]

Solve for [tex]v_{f3}[/tex]

[tex]a\:=\:\frac{v_f\:-\:v_i}{t}[/tex]

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

[tex](t)(-0.8\:m/s²)\:=\:-24\:m/s[/tex]

[tex]t\:=\:\frac{-24\:m/s}{-0.8\:m/s²}[/tex]

[tex]t\:=\:30\:s[/tex]

[tex]\\[/tex]

Solve for [tex]v_{ave3}[/tex]

[tex]v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}[/tex]

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

[tex]v_{ave}\:=\:12\:m/s[/tex]

[tex]\\[/tex]

Solve for [tex]d_3[/tex]

[tex]v\:=\:\frac{d}{t}[/tex]

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

[tex]360\:m\:=\:d[/tex]

[tex]d_3\:=\:360\:m[/tex]

[tex]\\[/tex]

Total Distance from Station A to Station B

[tex]d\:= \:d_1\:+\:d_2\:+\:d_3[/tex]

[tex]d\:= \:720\:m\:+\:36,000\:m\:+\:360\:m[/tex]

[tex]d\:= \:37,080\:m[/tex]

[tex]d\:= \:37.08\:km[/tex]

[tex]\\[/tex]

Final Answer:The distance between the two stations is 37.08 km

Related Questions

Corrections for curvature and refraction (c r) are applied to: ____________.
a. benchmark elevations (BM)
b. both foresight rod readings (FS) and backsight rod readings (BS)
c. foresight rod readings (FS)
d. backsight rod readings (BS)

Answers

Answer:

b.

Explanation:

Foresight rod reading is any measurement taken at a given sight to determine a particular elevation. The backsight rod readings are usually used to a point of certain elevation. It is added to the elevation to determine the height of the instrument. These are phenomena are used in differential leveling and are applied to corrections for curvature and refraction.

A new construction firm completed its first project of a residential building. But a few months later, the building developed cracks, and eventually the entire building crumbled to the ground. What caused the building to fall?
A. Poor air conditioning
B. Weak walls
C. External force
D. Furniture

Answers

It’s B bc if it has weak walls then it’s gonna fall

Answer:

it is weak walls

Explanation:

Which of these construction materials does the government restrict because of toxicity?
A.
lead
B.
silica
C.
concrete
D.
cement

Answers

Answer:

A

Explanation:

It is also a toxic material

Answer:

The correct answer is A. Lead.

Explanation:

I got it right on the Plato test.

(TCO 1) The cutoff frequency of a low pass filter is usually identified by its _____. Group of answer choices

Answers

Answer:

-3 dB point

Explanation:

A filter is a circuit deigned to pass or amplify certain frequencies while also preventing other frequencies from passing. There are different types of filter depending on the frequency range such as the low pass filter, high pass filter, bandstop filter, bandpass filter.

A low pass filter is a filter that allows the passage of frequencies below the cut off frequency and blocks frequencies above the cut off frequency. The cutoff frequency of a low pass filter is the frequency at the -3dB point.

The fan pressure differential gage on an air handler reads 12 cm H2O. What is this pressure differential in kiloPascals

Answers

Answer:

[tex]1.18\ \text{kPa}[/tex]

Explanation:

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

h = Height of reading = 12 cm

[tex]\rho[/tex] = Density of water = [tex]1000\ \text{kg/m}^3[/tex]

Pressure due to height difference is given by

[tex]P=\rho gh\\\Rightarrow P=1000\times 9.81\times 12\times 10^{-2}\\\Rightarrow P=1177.2\ \text{Pa}=1.1772\ \text{kPa}\approx 1.18\ \text{kPa}[/tex]

The pressure differential is [tex]1.18\ \text{kPa}[/tex].

The pressure differential will be "1.18 kPa".

According to the question,

Height of reading, [tex]h = 12 \ cm[/tex]Density of water, [tex]\rho = 1000 \ kg/m^3[/tex]Acceleration due to gravity, [tex]g = 9.8 \ m/s^2[/tex]

The formula will be:

→ [tex]P = \rho g h[/tex]

By putting the values, we get

      [tex]= 1000\times 9.8\times 12\times 10^{-2}[/tex]

      [tex]= 1177.2 \ Pa[/tex]

      [tex]= 1.18 \ kPa[/tex]

Thus the answer above is correct.

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which of the following activities can help expand engineers' creative thinking capabilities?
web design
team building exercises
journal keeping
trial and error

Answers

Answer:

team building exercises

Explanation:

Answer:   its not team building excercize

Explanation:

A certain power-supply filter produces an output with a ripple of 100 mV peak-to-peak and a dc value of 20 V. The ripple factor is

Answers

Answer: the ripple factor is 0.005

Explanation:

Given the data in the question;

we know that expression of ripple factor is;

r = Vr(pp) / Vdc

where Vr(pp) the peak to peak is ripple voltage ( 100mv = 0.1 V)

and Vdc is the dc value of the filter output voltage ( 20 V)

so we substitute our given values;

r = 0.1 / 20

r = 0.005

Therefore; the ripple factor is 0.005

The ripple factor of the given power supply filter is; γ = 0.005

We are given;

Ripple peak to peak voltage; V_r,pp = 100 mV = 0.1 V

DC value of the filter output voltage; V_dc = 20 V

Now, formula for the ripple factor is given as;

γ = V_rms/V_dc

Now, our ripple peak to peak voltage is also known as the rms value of ripple voltage at the output. Thus;

γ = 0.1/20

γ = 0.005

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What time ----–- the train arrve? ​​

Answers

I think it’s TIME FOR U TO GET A WATCH !!! lol I’m jk jk I believe it arrives at 12:33

The bulk density of a compacted soil specimen (Gs = 2.70) and its water content are 2060 kg/m^3 and 15.3%, respectively. If the specimen is soaked in a bucket of water for sev-eral days until it is fully saturated, what should the saturated density be?

Answers

Answer:

the saturated density should be

Explanation:

On a day when the barometer reads 755 mmhg, a tire pressure gage reads 204 kPa. The absolute pressure in the tire is:

Answers

Answer:

2.29mHg

Explanation:

We first do a conversion of gage pressure to mmHg

= 204(10³) / 13595(9.807)

= 204000/133326.165

= 1.53mHg

755 mmhg = 0.755

The pressure is therefore

P = p(atmosphere) + Of

= 1.53 + 0.755

= 2.29 mHg

This is the absolute pressure on this tire. So this answers this question.

The absolute pressure in the tire is equal to 2,285.13 mmHg.

Given the following data:

Atmospheric pressure of barometer = 755 mmHg.Gauge pressure of tire = 204 kPa.

To calculate the absolute pressure in the tire:

First of all, we would convert the value of gauge pressure in kPa to mmHg.

Conversion:

1 kPa = 7.50062 mmHg

204 kPa = [tex]204 \times 7.50062 = 1530.13\;mmHg[/tex]

The formula for absolute pressure.

Mathematically, absolute pressure is given by this formula:

[tex]P_{abs} = P_{atm} + P_{ga}[/tex]

Where:

[tex]P_{atm}[/tex] is the atmospheric pressure.[tex]P_{ga}[/tex] is the gauge pressure.

Substituting the parameters into the formula, we have;

[tex]P_{abs} = 755 + 1530.13[/tex]

Absolute pressure = 2,285.13 mmHg.

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One - tenth kilogram of air as an ideal gas with k= 1.4 executes a carnot refrigeration cycle as shown i fig. 5,16, the isothermal expansion occurs at - 23C with a heat transfer to the air of 3.4 kj. The isothermal compression occurs at 27C to a final volume of 0.01m. Using the results of prob. 5.80 adapted to the case, Determine (a) the pressure, in Kpa, at each of the four principal states (b) the work, in KJ for each of the four processes (c) the coefficient of performance

Answers

Answer:

Hello your question is incomplete attached below is the missing part

a) p1 = 454.83 kPa,  p2 = 283.359 Kpa , p3 = 536.423 kpa , p4 = 860.959kPa

b) W12 = 3.4 kJ, W23 = -3.5875 KJ, W34 = -4.0735 KJ, W41 = 3.5875 KJ

c) 5

Explanation:

Given data:

mass of air ( m ) = 1/10 kg

adiabatic index ( k ) = 1.4

temperature for isothermal expansion = 250K

rate of heat transfer ( Q12 ) = 3.4 KJ

temperature for Isothermal compression ( T4 ) = 300k

final volume ( V4 ) = 0.01m ^3

a)  Calculate the pressure, in Kpa, at each of the four principal states

from an ideal gas equation

P4V4 = mRT4 ( input values above )

hence P4 = 860.959kPa

attached below is the detailed solution

b) Calculate work done for each processes

attached below is the detailed solution

C) Calculate the coefficient of performance

attached below is detailed solution

The speed of sound is 1150 ft/s convert to mile/h

Answers

Answer:

784.090909mph

Explanation:

1ft/s=0.681818 mph

1150ft/s=0.681818 x 1150=784.090909

Convert 5m/h to yds/week

Answers

Rounded to a whole number, it would be 919 yds/week.

918.63517 is the answer.

DİGİTAL LOGİC DESİGN

Answers

Answer:

Uh- dude if you think I'm gonna download that, think twice...

not downloading that lol anyways haha

how much metal can be removed from a cracked drum to restore surface

Answers

There is a turn to limit stamped on it but as others have said, if it is cracked, you need to replace.

No amount of metal removal will restore a cracked surface, rather, there is a need to completely change the metal drum.

What is a metal drum?

This means a from of cylindal metal container that is used for shipping or storage of liquids.

When its surface is cracked, then, no amount of metal removal will restore a cracked surface, rather, there is a need to completely change the metal drum.

Read more about metal drum

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#SPJ2

Describe the differences between case hardening and through hardening, insofar as engineering applications of metals are concerned.

Answers

Answer:

The answer is below

Explanation:

Case hardening is a form of steel hardening that is applied on mild steel with a high temperature of heat.

It results in material forming a hard surface membrane, while the inner layer is soft.

It is mostly used for universal joints, construction cranes, machine tools, etc.

On the other hand, Through hardening is a form of steel hardening in engineering that involves heat treatment of carbon steel.

It increases the hardness and brittleness of the material.

It is often used for axles, blades, nuts and bolts, nails, etc.

Animation can occur before an action

Answers

Answer:Animation can occur before an action. Prepares the audience for the action. ... A pose or action should clearly communicate to the audience the attitude, mood, reaction or idea of the character. The effective use of long, medium or close up shots, as well as camera angles helps tell the story.

Explanation:May i have brainlist plz only if u wanna give me brainlist though have an nice day and stay safe.

ASAE 1060 Steel wire (1 mm diameter) is coated with copper to form a composite with a diameter of 2mm. Use the following properties for parts a, b, and c of question 2: The elastic modulus of copper is 110 GPa The yield stress of the copper is 140 MPa The coefficient of thermal expansion of the copper is 17 times 10^-6/degree C. The elastic modulus of steel is 205 GPa The yield stress of the steel is 280 MPa The coefficient of thermal expansion of the copper is 10 times 10^-6/degree C Determine: a. The elastic modulus of the composite b. The maximum force that the composite will carry before either material yields c. The coefficient of thermal expansion of the composite material.

Answers

Answer:

a) [tex]E_{m}[/tex] = 133.75 Gpa

b) Fnet = 560 N

c) thermal expansion of the composite material = 14.31 [tex]10^{-6 }[/tex] / °C

Explanation:

Solution:

a) Elastic Modulus of the composite:

Area of steel wire = [tex]\frac{\pi }{4}[/tex] x ([tex]0.001^{2}[/tex]) = 0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Area of Copper wire = [tex]\frac{\pi }{4}[/tex] x ([tex]0.002^{2}[/tex]) - 0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Area of Copper wire = 2.4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Young's Modulus of Composite mixture:

[tex]E_{m}[/tex] = [tex]F_{st}[/tex][tex]E_{st}[/tex] +  [tex]F_{Cu}[/tex][tex]E_{Cu}[/tex]     Equation 1

here,

[tex]F_{st}[/tex] = Stress in Steel

[tex]F_{Cu}[/tex] = Stress in Copper.

We know that,

F = P/A

F is inversely proportional to Area, so if area is large, stress will less and vice versa. So, Take

Ratio for area of steel = [tex]\frac{0.8. 10^{-6} }{(0.8 + 2.4) .10^{-6} }[/tex]

Ratio for area of steel = [tex]\frac{0.8}{3.2 }[/tex] = 0.25

Similarly, for Copper,

Ratio for area of copper = [tex]\frac{2.4. 10^{-6} }{(0.8 + 2.4) .10^{-6} }[/tex]

Ratio for area of copper = [tex]\frac{2.4 }{3.2}[/tex] = 0.75

Put these values in equation 1:

[tex]E_{m}[/tex] = [tex]F_{st}[/tex][tex]E_{st}[/tex] +  [tex]F_{Cu}[/tex][tex]E_{Cu}[/tex]    

[tex]E_{m}[/tex] = (0.25) [tex]E_{st}[/tex] +  (0.75)[tex]E_{Cu}[/tex]

We are given that,

  [tex]E_{st}[/tex] = 205 Gpa

[tex]E_{Cu}[/tex]  = 110 Gpa

So,

[tex]E_{m}[/tex] = (0.25) (205 Gpa) +  (0.75) (110 GPa)

[tex]E_{m}[/tex] = 51.25GPa + 82.5 Gpa

Hence, the Elastic Modulus of the composite will be:

[tex]E_{m}[/tex] = 133.75 Gpa

b) maximum force:

Fnet = Fst + Fcu

We know that F = (Yield Stress x Area)

F = fst x Ast + fcu x Acu

And we are given that,

Yield stress of Steel = 280 Mpa

Yield stress of Copper = 140 Mpa

And,

Ast = 0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Acu = 2.4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Just plugging in the values, we get:

F = (280 Mpa) (0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]) + (140 Mpa) (2.4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex])

F = 224 + 336

Fnet = 560 N    ( because Mpa = [tex]10^{6}[/tex] N/[tex]m^{2}[/tex])

So, it means the composite will carry the maximum force of 560N

c) Coefficient of Thermal Expansion:

Strain on both material is same upon loading so,

(ΔL/L)st = (ΔL/L)cu

by thermal expansion equation:

([tex]\alpha .[/tex]ΔT  + [tex]\frac{F}{A}[/tex][tex]. \frac{1}{Est}[/tex]) = [tex]\alpha .[/tex]ΔT  + [tex]\frac{F}{A}[/tex][tex]. \frac{1}{Ecu}[/tex])

Where [tex]\alpha[/tex] = Coefficient of Thermal expansion

Here, fst = -fcu = F

and ΔT = 1°

So,

Plugging in the values, we get.

( 10 x [tex]10^{-6}[/tex] x (1) + [tex]\frac{F}{0.8.10^{-6} } . \frac{1}{205 . 10^{9} }[/tex] ) = ( 17 x [tex]10^{-6}[/tex] x (1) + [tex]\frac{-F}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }[/tex] )

Solving for F, we get:

F = 0.71 N

Here,

fst = F = 0.71 N (Tension on Heating)

fcu = -F = 0.71 N ( Compression on Heating )

So, the combined thermal expansion of the composite material will be:

(ΔL/L)cu = ( 17 x [tex]10^{-6}[/tex] x (1°) + [tex]\frac{-0.71}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }[/tex] )

(ΔL/L)cu = ( 17 x [tex]10^{-6}[/tex] x (1°) - 2.69 x [tex]10^{-6}[/tex]

combined thermal expansion of the composite material = 14.31 [tex]10^{-6 }[/tex] / °C

The primary energy source for the controller in a typical control system is either brainlythe primary energy source for the controller in a typical control system is either

Answers

Answer:

a pneumatic or electric power

Explanation:

The primary energy source for the controller in a typical control system is either "a pneumatic or electric power."

This is because a typical control system has majorly four elements which include the following:

1. Sensor: this calculates the controlled variable

2. Controller: this receives and process inputs from the sensor to the controlled device as output

3. Controlled device: this tweak the controlled variable

4. Source of energy: this is the energy used to power the control system. It could be a pneumatic or electric power

At a high school science fair, Connor won first place for his replica of the Golden Gate Bridge. Connor liked the
project so much, he now wants to design bridges as a career. Which will best position Connor to do that?
earning a bachelor's degree in Civil Engineering from a four-year university, completing an internship, and
seeking a job at a private firm
earning a doctoral degree in Civil Engineering and seeking a job in the public works department of a state or
federal government
O completing an internship with a state or federal government before earning an associate degree in Civil
Engineering from a technical school
working at an entry-level job at a private engineering firm before earning a degree in Civil Engineering from a
four-year university

Answers

The correct answer is A. Earning a bachelor's degree in Civil Engineering from a four-year university, completing an internship, and seeking a job at a private firm.

Explanation:

In the U.S. and many countries, the best to start a career is to enroll in a formal educational program at a university or college. This helps students learn concepts, theories, methods, etc. they need for their profession. Moreover, a degree such as a bachelor's degree is required by employers. In this context, the first step for Connor is to earn a bachelor's degree in Civil Engineering.

Besides this, an internship is recommended after earning a degree because this is the way students can gain real-life work experience, which is considered positive by employers. This means the next step should be an internship.

Finally, Connor can seek a job to design bridges and other buildings because after the degree and internship he will have the experience and knowledge required by employers and by the job.

Technician A says that tailor-rolled parts may be used for collision energy managements.

Technician B says that tailor-welded parts are aluminum and steel parts joined together. Who is right?


A Only

B only

Both A and B

Neither A nor B

Answers

The correct answer to your problem is the answers of a and b

A 30*30 cm cross section concrete pile with length of 18 m, the carrying capacity of the pile at
the pile tip is 9445 KN/m2 and the average unit skin resistance is 25 kN/m2. The ultimate load
carrying capacity of this pile is:
Select one:
0 2231 KN
2111 KN
1390 KN
1300 KN​

Answers

Answer:

b

Explanation:

Trust me. I am not the smartest though...

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