1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
nadezda [96]
3 years ago
8

A train consists of a 50 Mg engine and three cars, each having a mass of 30 Mg . If it takes 75 s for the train to increase its

speed uniformly to 45 km/h , starting from rest, determine the force T, developed at the coupling between the engine E and the first car A. The wheels of the engine provide a resultant frictional tractive force F which gives the train forward motion, whereas the car wheels roll freely. Also, determine F acting on the engine wheels.
Engineering
1 answer:
ohaa [14]3 years ago
7 0

Answer:

T = 15 kN

F = 23.33 kN

Explanation:

Given the data in the question,

We apply the impulse momentum principle on the total system,

mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

we substitute

[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ ×  ( 45 × 1000 / 3600 )  

F( 75 - 0 ) =  1.75 × 10⁶

The resultant frictional tractive force F is will then be;

F =  1.75 × 10⁶ / 75

F = 23333.33 N

F = 23.33 kN

Applying the impulse momentum principle on the three cars;

mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ ×  ( 45 × 1000 / 3600 )  

F(75-0) = 1.125 × 10⁶

The force T developed is then;

T =  1.125 × 10⁶ / 75

T = 15000 N

T = 15 kN

You might be interested in
How do technological artifacts affect the way that you live?
Maslowich

Answer:

Artefacts can influence our actions in several ways. They can be instruments, enabling and facilitating actions, where their presence affects the number and quality of the options for action available to us. They can also influence our actions in a morally more salient way, where their presence changes the likelihood that we will actually perform certain actions. Both kinds of influences are closely related, yet accounts of how they work have been developed largely independently, within different conceptual frameworks and for different purposes. In this paper I account for both kinds of influences within a single framework. Specifically, I develop a descriptive account of how the presence of artefacts affects what we actually do, which is based on a framework commonly used for normative investigations into how the presence of artefacts affects what we can do. This account describes the influence of artefacts on what we actually do in terms of the way facts about those artefacts alter our reasons for action. In developing this account, I will build on Dancy’s (2000a) account of practical reasoning. I will compare my account with two alternatives, those of Latour and Verbeek, and show how my account suggests a specification of their respective key concepts of prescription and invitation. Furthermore, I argue that my account helps us in analysing why the presence of artefacts sometimes fails to influence our actions, contrary to designer expectations or intentions.

When it comes to affecting human actions, it seems artefacts can play two roles. In their first role they can enable or facilitate human actions. Here, the presence of artefacts changes the number and quality of the options for action available to us.Footnote1 For example, their presence makes it possible for us to do things that we would not otherwise be able to do, and thereby adopt new goals, or helps us to do things we would otherwise be able to do, but in more time, with greater effort, etc

Explanation:

Technological artifacts are in general characterized narrowly as material objects made by (human) agents as means to achieve practical ends. ... Unintended by-products of making (e.g. sawdust) or of experiments (e.g. false positives in medical diagnostic tests) are not artifacts for Hilpinen.

3 0
3 years ago
The water in a 25-m-deep reservoir is kept inside by a 140-m-wide wall whose cross section is an equilateral triangle as shown i
koban [17]

Answer:  (a) 9.00 Mega Newtons or 9.00 * 10^6 N

               (b)  17.1 m

Explanation:  The length of wall under the surface can be given by

                                            b=25m/sin(60)\\=28.867

The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.

F(resultant) = Pavg ( A) = (Patm +  \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N

Noting from the Bernoulli  equation that

Po/\rho g sin60 = 100000/1000 * 9.81* sin(60) = 11.77 m \\ \\

From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:

Yp = s+\frac{b}{2} +\frac{b^2}{s+\frac{b}{2}+Po/\rho g sin60}= 0+\frac{28.87}{2} +\frac{28.87^2}{0+\frac{28.87}{2}+100000 /1000 *9.81 sin60} = 17.1 m

Substituting the values gives us the the distance of the surface to be equal to = 17.1 m

7 0
3 years ago
Don't break or crush mercury-containing lamps because mercury powder may be released.
alexandr402 [8]
A.

It would be released without a doubt so be careful!

Hope this helps :)
5 0
3 years ago
Name three types of large motor vehicles with which you might share the roadway. Explain how you can reduce risk when interactin
alexdok [17]

Answer:

rucks, buses, RVs, trolleys

Explanation:

What are some different types of vehicles that we share the road with?

Driving is a complex task, and for safe driving you need to know not just the rules of sharing the road with other cars, but with variety of other types of vehicles: trucks, buses, RVs, trolleys, motorcycles, bicycles and of course pedestrians.

8 0
2 years ago
Flank wear data were collected in a series of turning tests using a coated carbide tool on hardened alloy steel at a feed of 0.3
Paladinen [302]

Answer:

A) n =  0.6143, c ≈ 640m/min

B) n = 0.6143 , c = 637.53m/min

Explanation:

using the given data

A) A plot of flank wear as a function of time and also A plot for tool when

Flank wear is 0.75 and cutting edge speed is 100m/min, Time of cutting edge is said to be 20.4 min  also for cutting edge speed of 155m/min , time for cutting edge is 10 min

is attached below

calculate for the constant N from the second plot

note : the slope will be negative because cutting speed decreases as time of cutting increase

V1 = 100m/min , V2 = 155m/min,  T1 = 20.4 min, T2 = 10 min

= - N = \frac{In(V2) - In(V1)}{In(T2)-ln(T1)}

therefore  - N = \frac{5.043 - 4.605}{2.302 -3.015}

                       = - 0.6143

THEREFORE  ( N ) = 0.6143

Determine for the constant C from the second plot as well

note : C is the intercept on the cutting speed axis in 1 min tool life

connecting the two points with a line and extend it to touch the cutting speed axis and measure the value at that point

hence   C ≈ 640m/min

B) Calculate the values of  N and C in the Taylor equation solving simultaneous equations

using the above cutting speed and time of cutting values we can find the constant N via Taylor tool life equation

Taylor tool life equation = vT = C ------------- equation 1

cutting speed = v = 100m/min and 155m/min

tool life = T = 20.4 min and 10 min

also constant  n and c are obtained from the previous plot

back to taylor tool life equation = 100 * 20.4 = C

therefore C = (100)(20.4)^n  ---------------- equation 2

also using the second values of  v and T

taylor tool life equation = 155 * 10 = C

therefore C = ( 155 )(10)^n ----------------- equation 3

Equate equation 2 and equation 3 and solve simultaneously

(100)(20.4)^n = (155)(10)^n

To find N

take natural log of both sides of the equation

= In ((100)(20.4)^n) = In((155)(10)^n)

= In (100) + nIn(20.4) = In(155) + nIn(10)^n

= n(3.0155) - n (2.3026) = 5.043 - 4.605

= 0.7129 n = 0.438

therefore n = 0.6143

To find C

substitute 0.6143 for n in equation 2

C = (100)(20.4) ^ 0.6143

C = 637.53 m/min

Attached are the two plots for solution A

7 0
3 years ago
Other questions:
  • The popularity of orange juice, especially as a breakfast drink, makes it an important factor in the economy of orange-growing r
    14·1 answer
  • When a mesh in a circuit contains an independent or dependent current source, this leads to a special case of mesh-current analy
    14·1 answer
  • Which branch of engineering studies the physical behavior of metallic elements?
    8·2 answers
  • The unit for volume flow rate is gallons per minute, but cubic feet per second is preferred. Use the conversion factor tables in
    13·1 answer
  • Sam promises to pay Sandy $2,000 in four years and another $3,000 four years later for a loan of $2,000 from Sandy today. What i
    8·1 answer
  • How do we need to prepare for the future?
    10·1 answer
  • The wave-particle duality theory is the first adequate explanation of which one of the following observations about the hydrogen
    10·1 answer
  • How does manufacturing affect the economy and society?
    14·2 answers
  • A two-bus power system is interconnected by one transmission line. Bus 1 is a generator bus with specified terminal voltage magn
    6·1 answer
  • 1. Band saw lower wheel does not require a guard * true or false 2. Band saw upper guide should be adjusted to within 1/8" of th
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!