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Vlad1618 [11]
2 years ago
15

What is a splitter gearbox​

Engineering
1 answer:
OLga [1]2 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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Please help me with this question
Nat2105 [25]

Answer:

The answer is c and the teacher helped me

Explanation:

i had help from the tescger and the assignment is done

5 0
1 year ago
The yield of a chemical process is being studied.The two most important variables are thought to be the pressure and the tempera
anyanavicka [17]

Answer:

<u>note:</u>

<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>

Download docx
7 0
3 years ago
Using any of the bilinear transform, matched pole-zero, or impulse invariance techniques in converting a continuous-time system
leonid [27]

Answer:

A. True

The bilinear transform is employed in digital signal processing and discrete-time control theory which helps in transforming continuous-time system representations to discrete-time

4 0
3 years ago
Read 2 more answers
Bulk wind shear is calculated by finding the vector difference between the winds at two different heights. Using the supercell w
ivanzaharov [21]

Answer:

See explaination

Explanation:

2. 0-1 km shear value: taking winds at 1000mb and 850 mb

15 kts south easterly and 50 kts southerly

Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts

3. 0-6 km shear value: taking winds at 1000 mb and 500 mb

15 kts south easterly and 40 kts westerly

Vector difference 135/15 and 270/40 will be 281/51 kts

please see attachment

5 0
3 years ago
A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate
kherson [118]

Explanation:

thermal expansion ∝L = (δL/δT)÷L ----(1)

δL = L∝L + δT ----(2)

we have δL = 12.5x10⁻⁶

length l = 200mm

δT = 115°c - 15°c = 100°c

putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

= 0.25 MM

L₂ = L + δ L

= 200 + 0.25

L₂ = 200.25mm

12.5X10⁻⁶ *115-15 * 20

= 0.025

20 +0.025

D₂ = 20.025

as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0

3 0
3 years ago
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