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2 years ago

11

How to clean a snowblower carburetor without removing it.

1 answer:

Anarel [89]2 years ago

8 0

Answer:

Spray carburetor cleaner on the inside of the bowl and wipe the liquid, dirt, and concentrated fuel off of it. Now take the main jet, spray the cleaner through it and wipe off the dirt. Then take a copper wire, scrub it through the tiny holes in the jet to complete the cleaning process.

Explanation:

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A shaft is made of an aluminum alloy having an allowable shear stress of τallow = 100 MPa. If the diameter of the shaft is 100 m

Elanso [62]

Answer:

A shaft is made of an aluminum alloy having an allowable shear stress of T_allow = 100 MPa. If the diameter of the shaft is 100 mm, determine the maximum torque T that can be transmitted.

Explanation:

8 0

3 years ago

Read 2 more answers

If a fuel line is routed through a compartment parallel with an electrical wire bundle, the fuel line should be installed ______

Gala2k [10]

Below so if it was to leak or bust it wouldn’t be much of a harm to anyone

8 0

3 years ago

WARNING:<br><br> when people put links in the answer it is a virus DO NOT DOWNLOAD IT

kobusy [5.1K]

Answer:

Thank you for this!

Explanation:

I was about to click it on a question I saw.

6 0

2 years ago

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In a cellular phone system, a mobile phone must be paged to receive a phone call. However, paging attempts don’t always succeed

Alina [70]

Answer:

The correct response will be "0.992". The further explanation to the following question is given below.

Explanation:

The probability that paging would be beneficial becomes 0.8

Effective paging at the very first attempted is 0.8

On the second attempt the success probability will be:

⇒ $0.2\times 0.8$

⇒ $0.16$

On the third attempt the success probability will be:

⇒ $0.2\times 0.2\times 0.8$

⇒ $0.032$

So that the success probability will be:

⇒ $0.8 + 0.16 + 0.032$

⇒ $0.992$

7 0

3 years ago

Six forces act on a beam that forms part of a building's

IrinaVladis [17]

Answer:

<h2> FA = 13 kN </h2><h2>FG = 15.3 kN</h2>

Explanation:

write each force in terms of magnitude and directions

Fx = F sin Ф

Fy = F cos Ф

where Ф is to be measured from x axis.

∑F at y = o

FAy + FBy + FCy + FDy + FEy + FGy = 0

∑F at x = o

FAx + FBx + FCx + FDx + FEx + FGx = 0

Let

FA = FA sin (110) + FA cos (110)

FB = 20 sin (270) + 20 cos (270)

FC = 16 sin (140) + 16 cos (140)

FD = 9 sin (40) + 9 cos (40)

FE = 20 sin (270) + 20 cos (270)

FG = FG sin (50) + FG cos (50)

add x and y forces:

FAx + FBx + FCx + FDx + FEx + FGx = 0

FAy + FBy + FCy + FDy + FEy + FGy = 0

FA sin (110) + 0 + 16 sin (140) + 9 sin (40) + 0 + FG sin (50) = 0

FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0

FA sin (110) + 0 + 10.285 + 5.785 + 0 + FG sin (50) = 0

FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0

FA sin (110) + 16.070 + FG sin (50) = 0

FA cos (110) - 45.363 + FG cos (50) = 0

solving for FA, and FG

FA = 13 kN

FG = 15.3 kN

7 0

3 years ago