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Kisachek [45]
3 years ago
5

when rolling up the landing gear after coupling, you should use _____ gear until the tractor is supporting the weight of the tra

iler.
Engineering
1 answer:
musickatia [10]3 years ago
6 0

Coupling is a process through which a tractor and other farm equipment are coupled (joined) together through the use of a pickup hitch and a tongue, while using them to work.

Basically, a coupling assembly connects (joins or couples) a towed equipment such as a trailer having a tongue to a tractor that has a pickup hitch.

Some of the precautions which must be taken before, during and after coupling a tractor to a towed equipment:

  • You should check your path for any hazard before you begin coupling.
  • You should check the normal air pressure for the braking system of the towed equipment.
  • You should use a<u> low gear</u> when rolling up the landing gear after coupling the two equipment. This is to ensure that the rate at which the landing gear is moving isn't too fast considering the weight of the trailer.

Read more: brainly.com/question/18270863?referrer=searchResults

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Answer: 5mW

Explanation:

8 0
3 years ago
A helical compression spring is made with oil-tempered wire with wire diameter of 0.2 in, mean coil diameter of 2 in, a total of
Naya [18.7K]

Answer:

a. Solid length Ls = 2.6 in

b. Force necessary for deflection Fs = 67.2Ibf

Factor of safety FOS = 2.04

Explanation:

Given details

Oil-tempered wire,

d = 0.2 in,

D = 2 in,

n = 12 coils,

Lo = 5 in

(a) Find the solid length

Ls = d (n + 1)

= 0.2(12 + 1) = 2.6 in Ans

(b) Find the force necessary to deflect the spring to its solid length.

N = n - 2 = 12 - 2 = 10 coils

Take G = 11.2 Mpsi

K = (d^4*G)/(8D^3N)

K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in

Fs = k*Ys = k (Lo - Ls )

= 28(5 - 2.6) = 67.2 lbf Ans.

c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.

For C = D/d = 2/0.2 = 10

Kb = (4C + 2)/(4C - 3)

= (4*10 + 2)/(4*10 - 3) = 1.135

Tau ts = Kb {(8FD)/(Πd^3)}

= 1.135 {(8*67.2*2)/(Π*2^3)}

= 48.56 * 10^6 psi

Let m = 0.187,

A = 147 kpsi.inm^3

Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi

Ssy = 0.50 Sut

= 0.50(198.6) = 99.3 kpsi

FOS = Ssy/ts

= 99.3/48.56 = 2.04 Ans.

7 0
4 years ago
4. If a hot wire is shorted to ground, what will usually happen?
barxatty [35]

Answer:

nothing will happen

Explanation:

except for whatever that wire feeds will lose power

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5 0
2 years ago
Write a program with total change amount as an integer input, and
AURORKA [14]

Answer:

amount = int(input())

#Check if input is less than 1

if amount<=0:

    print("No change")

else: #If otherwise

    #Convert amount to various coins

    dollar = int(amount/100) #Convert to dollar

    amount = amount % 100 #Get remainder after conversion

    quarter = int(amount/25) #Convert to quarters

    amount = amount % 25 #Get remainder after conversion

    dime = int(amount/10) #Convert to dimes

    amount = amount % 10 #Get remainder after conversion

    nickel = int(amount/5) #Convert to nickel

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    #Print results

    if dollar >= 1:

          if dollar == 1:

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          else:

                print(str(dollar)+" Dollars")

    if quarter >= 1:

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          else:

                print(str(quarter)+" Quarters")

    if dime >= 1:

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          else:

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    if nickel >= 1:

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          else:

                print(str(nickel)+" Nickels")

    if penny >= 1:

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Explanation:

8 0
3 years ago
A 4,000-km^2 watershed receives 102cm of precipitation in one
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Answer:

1) The change in storage of the catchment is 707676800 cubic meters.

2) The runoff coefficient of the catchment is 0.83.

Explanation:

The water budget equation of the catchment can be written as

P+Q_{in}=ET+\Delta Storage+Q_{out}+I

where

'P' is volume of  precipitation in the catchment =Area\times Precipitation

Q_{in} Is the water inflow

ET is loss of water due to evapo-transpiration

\Delta Storage is the change in storage of the catchment

Q_{out} is the outflow from the catchment

I is losses due to infiltration

Applying the values in the above equation and using the values on yearly basis (Time scale is taken as 1 year) we get

4000\times 10^{6}\times 1.02+0=0.40\times 4000\times 10^{6}+\Delta Storage+34.2\times 3600\times 24\times 365\times 5.5\times 10^{-9}\times 4000\times 10^{6}\times 3600\times 24\times 365

\therefore \Delta Storage=707676800m^3

Part b)

The runoff coefficient  C is determined as

C=\frac{P-I}{P}

where symbols have the usual meaning as explained earlier

\therefore C=\frac{102-5.5\times 10^{-7}\times 3600\times 24\times 365}{102}=0.83

5 0
3 years ago
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