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3 years ago

4. Which of the following is the first thing you should do when attempting

Engineering

2 answers:

Hoochie [10]3 years ago

6 0

hey there! i am not really good at this topic, but I’m pretty sure the answer is C.

let me know if you get it right! :)

Klio2033 [76]3 years ago

3 0

Answer: B. Turning on your hazard lights.

Explanation:

Because...

that indicates the drive behind you to go in front of you and indicator lights that flash in unison to warn other drivers that the vehicle is a temporary obstruction.

* Hopefully this helps:) Mark me the brainliest:)!!!

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What are the disadvantages of having a liquid cooled engine?

Feliz [49]

One notable disadvantage of liquid cooling over air cooling is that it is considerably costly to set up. Cooling fans are prevalent in the market, and this overabundance of supply means they are cheap. The components of a liquid cooling system can be expensive.

5 0

2 years ago

Two thousand pieces will flow through from the first machine A to the final machine F based on the given sequence of operations.

Vlad1618 [11]

The total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.

"Your question is not complete, it seems to be missing the following information;"

The sequence of operation is A - E - D - C - B - A - F

The given parameters;

number of pieces that will flow from the first machine A to machine F, = 2,000 pieces
initial unit load specified in the first machine, L₁ = 50
final unit load, L₂ = 100
the capacity of the vehicle = 1 unit load

The given sequence of operation of the vehicle;

A - E - D - C - B - A - F

the vehicle makes 6 trips for 100 unit loads

The total number of trips that the vehicle has to make, in order to transport the 2000 pieces of the load given, is calculated as follows.

100 unit loads ----------------- 6 trips

2000 unit loads --------------- ?

$= \frac{2000}{100} \times 6\\\\= 120 \ trips$

Thus, the total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.

Learn more here:brainly.com/question/21468592

6 0

2 years ago

What type of engineering do you think would help solve this SDG???

OleMash [197]

Answer:

Explanation:

Planning engineering

4 0

3 years ago

Find the following for an input of 120 VAC(RMS), 60 hertz, given a 10:1 stepdown transformer, and a full-wave bridge rectifier.

atroni [7]

Answer:

(i) 169.68 volt

(ii) 16.90 volt

(iii) 16.90 volt

(iv) 108.07 volt

(v) 2.161 A

Explanation:

Turn ratio is given as 10:1

We have given that input voltage $v_p=120volt$

(i) We know that peak voltage is give by $v_{peak}=\sqrt{2}v_p=\sqrt{2}\times 120=169.68volt$

(ii) We know that for transformer $\frac{v_p}{v_s}=\frac{n_p}{n_s}$

So $\frac{169.08}{v_s}=\frac{10}{1}$

$v_s=16.90volt$

So peak voltage in secondary will be 16.90 volt

(iii) Peak voltage of the rectifier will be equal to the peak voltage of the secondary

So peak voltage of the rectifier will be 16.90 volt

(iv) Dc voltage of the rectifier is given by $v_{dc}=\frac{2v_m}{\pi }=\frac{2\times 1.414\times 120}{3.14}=108.07volt$

(v) Now dc current is given by $i_{dc}=\frac{v_{dc}}{R}=\frac{108.07}{50}=2.1614A$

4 0

3 years ago

Consider an 8-car caravan, where the propagation speed is 100 km/hour, each car takes 1 minute to pass a toll both. The caravan

melamori03 [73]

Answer:

A. 36 minutes

B. 120 minutes

i. 144 minutes

ii. 984 minutes

D. Car 1 is 1.67km ahead of Cat 2 when Car 2 passed the toll B.

E. 98.33km

Explanation

Given

dAb = 10km

dBc = 10km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 0.2 hours

= 24 minutes + 0.2 * 60 minutes

= 24 minutes + 12 minutes

= 36 minutes

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 0.2 hours

= 3 minutes + 0.2 * 60 minutes

= 3 minutes + 12 minutes

= 15 minutes

Total End delay for 8 cars = 8 * 15 = 120 minutes

Given

dAb = 100km

dBc = 100km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

i. Cars travelling together

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 2 hours

= 24 minutes + 2 * 60 minutes

= 24 minutes + 120 minutes

= 144 minutes

ii. Cars travelling separately

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 2 hours

= 3 minutes + 2 * 60 minutes

= 3 minutes + 120 minutes

= 123 minutes

Total End delay for 8 cars = 8 * 123 = 984 minutes

Distance = 100km

Time = 1 min/car

Car 1 is 1 minute ahead of car 2 --- at toll A and B

If car 1 leaves toll B after 10 minutes then cat 2 leaves after 11 minutes

Time delay = 11 - 10 = 1 minute

Distance = time * speed

= 1 minute * 100km/hr

= 1 hr/60 * 100 km/hr

= 100/60

= 1.67km

Given

Distance = 100km

Distance behind = 1.67

Maximum value of dBc = 100km - 1.67km = 98.33km

The maximum distance that can be reached is 98.33km

7 0

3 years ago