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Vlad1618 [11]
3 years ago
15

What is a splitter gearbox​

Engineering
1 answer:
OLga [1]3 years ago
6 0

Answer:

The splitter transmission also uses a high-low division. Instead of having a high and low section, the gears are split into two so that each position of the gear shift is used for two gears (one high and one low).

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¿Qué es la masonería?​
timofeeve [1]

Answer: freemasonry is Being a Mason is about a father helping his son make better decisions; a business leader striving to bring morality to the workplace; a thoughtful man learning to work through tough issues in his life.

Explanation:

3 0
3 years ago
When buttons or switches are pressed by humans for arbitrary periods of time, we need to convert a signal level to a pulse. In t
ddd [48]

Answer:

The FSM uses the states along with the generation at the P output on each of the positive edges of the CLK. The memory stores the previous state in the machine and the decoder generates a P output based on the previous state.

Explanation:

The code is in the image.

6 0
3 years ago
Water is flowing at a rate of 0.15 ft3/s in a 6 inch diameter pipe. The water then goes through a sudden contraction to a 2 inch
Georgia [21]

Answer:

Head loss=0.00366 ft

Explanation:

Given :Water flow rate Q=0.15 \frac{ft^{3}}{sec}

         D_{1}= 6 inch=0.5 ft

        D_{2}=2 inch=0.1667 ft

As we know that Q=AV

A_{1}\times V_{1}=A_{2}\times V_{2}

So V_{2}=\frac{Q}{A_2}

     V_{2}=\dfrac{.015}{\frac{3.14}{4}\times 0.1667^{2}}

     V_{2=0.687 ft/sec

We know that Head loss due to sudden contraction

           h_{l}=K\frac{V_{2}^2}{2g}

If nothing is given then take K=0.5

So head lossh_{l}=(0.5)\frac{{0.687}^2}{2\times 32.18}

                                    =0.00366 ft

So head loss=0.00366 ft

4 0
3 years ago
To be able to solve problems involving force, moment, velocity, and time by applying the principle of impulse and momentum to ri
coldgirl [10]

Answer:

see explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

The attached files has the solved problem.

3 0
3 years ago
Read 2 more answers
The extruder head in a fused- deposition modeling setup has a diameter of 1.25 mm (0.05 in) and produces layers that are 0.25mm
Angelina_Jolie [31]

Answer:

The time taken will be "1 hour 51 min". The further explanation is given below.

Explanation:

The given values are:

Number of required layers:

= \frac{38}{0.25}

= 152 \ layers

Diameter (d):

= 1.25 mm

Velocity (v):

= 40 mm/s

Now,

The area of one layer will be:

= 38\times 38 \ mm^2

= 1444 \ mm^2

The area covered every \second will be:

= d\times v

= 1.25\times 40

= 50 \ mm^2

The time required to deposit one layer will be:

= \frac{1444}{50}

= 28.88 \ sec

The time required for one layer will be:

= 15 \ sec

∴ Total times required for one layer will be:

= 15+28.88

= 43.88 \ sec

So,

Number of layers = 152

Therefore,

Total time will be:

= 152\times 43.88

= 6669.76 \ sec

= 1 \ hour \ 51 \ min

6 0
4 years ago
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