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Dahasolnce [82]
3 years ago
13

A front wheel drive vehicle with four wheel disc brakes is pulling to the left. Tech A says an external kink or internal restric

tion in the LF brake line will result in this condition. Tech B says to use a compression fitting to repair a section of brake line. Who is correct? Tech A Tech A Tech B Tech B Both Both Neither
Engineering
1 answer:
STALIN [3.7K]3 years ago
5 0

Answer:

Tech A is correct.          

Explanation:

A front-wheel-drive pulling to the left can result from several factors. One of them is definitely a faulty break.

A correct diagnosis linking the problem to the brakes is when there is an internal restriction and the pull is constant to one side and gets worse when the brakes are applied.

To confirm this, one would need to lift the vehicle and rotate each wheel by hand to check for excessive friction.

So the restriction may be caused by:

  • brake calipers that are sticky to the drum
  • too much brake fluid in the brake master cylinder - this prevents the caliper pistons from retracting when the brakes are released
  • misadjusted drum brakes and or parking brakes.

Cheers

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Propane gas C3H8 enters a combustion chamber operating at steady state condition at 1 bar, 25ºC and is burned with 150% theoreti
GalinKa [24]

Answer:

see explaination

Explanation:

Balanced equation or stoichiometry equation means in a product after reaction there is no unburned carbon compound left or we can say the oxygen is sufficient to combine with all the carbon and hydrogen moleculs to form Carbon-dioxide and water respectively.

The dew point temperature of balanced equation will be 100°c because water vapour bis present in it and it will condense at 100°c at 1 bar pressure while the other products need much lower temperatures to liquify.

See attachment.

6 0
3 years ago
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According to Moore and Marra's (2005) case study, which observed two online courses, students in the first course implemented a
Archy [21]

Your question is incomplete, please let me assume this to be your complete question;

ORIGINAL SOURCE:

When instructors are creating discussion board activities for online courses, at least two questions must be answered. First, what is the objective of the discussions? Different objectives might be to create a "social presence" among students so that they do not feel isolated, to ask questions regarding assignments or topics, or to determine if students understand a topic by having them analyze and evaluate contextual situations. Based on the response to this question, different rules might be implemented to focus on the quality of the interaction more so than the quantity. The second question is, how important is online discussions in comparison to the other activities that students will perform? This question alludes to the amount of participation that instructors expect from students in online discussions along with the other required activities for the course. If a small percentage of student effort is designated for class participation, our results show that it can affect the quality and quantity of interactions.

References:

Moore, J. L., & Marra, R. M. (2005) A comparative analysis of online discussion participation protocols.Journal of Research on Technology in Education, 38(2), 191-212.

STUDENT VERSION:

According to Moore and Marra's (2005) case study, which observed two online courses, students in the first course implemented a constructive argumentation approach while students in second course had less structure for their postings. As they stated, when instructors create online discussion board activities, they must answer at least two questions. These questions are: "What is the objective of the discussions?" And "How important are online discussions in comparison to the other activities that students will perform?". According to their findings, the discussion activities that were designed based on the answers to these questions can influence the quality and quantity of interactions (Moore & Marra, 2005).

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Moore, J. L., & Marra, R. M. (2005) A comparative analysis of online discussion participation protocols.Journal of Research on Technology in Education, 38(2), 191-212.

Which of the following is true for the students work;

Word-for-word plagiarism

Paraphrasing plagiarism

Not Plagiarism

ANSWER: IT IS NOT PLAGIARISM

Explanation: plagiarism is the act of extracting knowledge from someone's literature work, without acknowledging the literature work. In other words, this can be called a theft of knowledge, because when you failed to acknowledge the literature source that helped you to produce your paper work, it means you have claimed to be the original owner of that knowledge.

This is not a Plagiarism because the student has acknowledged the source of the knowledge in it's literature work. The original source and the student has cited the same literature work, this why their work looks similar but not exactly the same. So the student has not committed Plagiarism

4 0
4 years ago
For better thermal control it is common to make catalytic reactors that have many tubes packed with catalysts inside a larger sh
Paul [167]

Answer:

the  pressure drop  is 0.21159 atm

Explanation:

Given that:

length of the reactor L = 2.5 m

inside diameter of the reactor d= 0.025 m

diameter of alumina sphere dp= 0.003 m

particle density  = 1300 kg/m³

the bed void fraction \in =  0.38

superficial mass flux m = 4684 kg/m²hr

The Feed is  methane with pressure P = 5 bar and temperature T = 400 K

Density of the methane gas \rho = 0.15 mol/dm ⁻³

viscosity of methane gas \mu = 1.429  x 10⁻⁵ Pas

The objective is to determine the pressure drop.

Let first convert the Density of the methane gas from 0.15 mol/dm ⁻³  to kg/m³

SO; we have :

Density =  0.15 mol/dm ⁻³  

Molar mass of methane gas (CH₄) = (12 + (1×4) ) = 16 mol

Density =  0.1 5 *\dfrac{16}{0.1^3}

Density =  2400

Density \rho_f =  2.4 kg/m³

Density = mass /volume

Thus;

Volume = mass/density

Volume of the methane gas =  4684 kg/m²hr / 2.4 kg/m³

Volume of the methane gas = 1951.666 m/hr

To m/sec; we have :

Volume of the methane gas = 1951.666 * 1/3600 m/sec = 0.542130 m/sec

Re = \dfrac{dV \rho}{\mu}

Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}

Re=2276.317705

For Re > 1000

\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}

\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}

\Delta P=8575.755212*2.5

\Delta = 21439.38803 \ Pa

To atm ; we have

\Delta P = \dfrac{21439.38803 }{101325}

\Delta P =0.2115903087  \ atm

ΔP  ≅  0.21159 atm

Thus; the  pressure drop  is 0.21159 atm

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