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3 years ago

How does a carburetor using two diaphragms works to supply fuel to the engine.

Engineering

1 answer:

Tems11 [23]3 years ago

5 0

Answer: well me helping my dad with cars

A flexible diaphragm forms one side of the fuel chamber and is arranged so that as fuel is drawn out into the engine, the diaphragm is forced inward by ambient air pressure. ... As fuel is replenished the diaphragm moves out due to fuel pressure and a small spring, closing the needle valve.

Explanation:

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Since the application can be the prospective employers first impression, it is important to a. always answer questions truthfull

loris [4]

Answer:

Definitely D

Explanation:

You should do all of those things in an application.

7 0

4 years ago

Read 2 more answers

This is an essential safety procedure that protects workers from injury while working on or near electrical circuits and equipme

Mice21 [21]

Answer:

Option B

Lockout/ Tagout

Explanation:

lockout and tagout is a safety procedure used in most industries and work places. This ensures that during any maintenance work, that dangerous machines are properly shut off and not able to be started up again prior to the completion of maintenance or repair work.

The procedure includes:

1. Turns off and disconnects the machinery or equipment from its energy source(s) before performing service or maintenance.

2. The authorized employee(s) either lock or tag the energy-isolating device(s) to prevent the release of hazardous energy.

8 0

4 years ago

MODIFIED-BOTTOM-UP-CUT-ROD(p, n, c) to return not only the value but the actual solution, too. Hint: It is similar to how array

Vaselesa [24]

Answer:

Matrix chain multiplication

M[i,j] = M[i,k] + M[(k+1),j] + p[i-1]*p[k]*p[j] i<=k<j

p[] = {5,10,3,12,5,50}

M[0][0] = 0,M[1][1] = 0,M[2][2] = 0,M[3][3] = 0,M[4][4] = 0,M[5][5] = 0,

M[1][2] = M[1][1]+M[2][2]+p[0]*p[1]*p[2] = 0+0+5*10*3 = 150

M[2][3] = M[3][3]+M[2][2]+p[1]*p[2]*p[3] = 0+0+10*3*12 = 360

M[3][4] = M[3][3]+M[4][4]+p[2]*p[3]*p[4] = 0+0+3*12*5 = 180

M[4][5] = M[4][4]+M[5][5]+p[3]*p[4]*p[5] = 0+0+12*5*50 = 3000

M[1][3] = min{M[1][1]+M[2][3]+p[0]*p[1]*p[3] , M[1][2]+M[3][3]+p[0]*p[2]*p[3]}

= {0 + 360 + 600 , 150+0+180} = {960,330} = 330

M[2][4] = min{M[2][2]+M[3][4]+p[1]*p[2]*p[4] , M[2][3]+M[4][4]+p[1]*p[3]*p[4]}

= {0 + 180 + 150 , 360+0+600} = {960,330} = 330

M[3][5] = min{M[3][3]+M[4][5]+p[2]*p[3]*p[5] , M[3][4]+M[5][5]+p[2]*p[4]*p[5]}

= {0 + 3000 + 1800 , 180+0+750} = {4800,930} = 930

M[1][4] = min{M[1][1] + M[2][4] +p[0]*p[1]*p[4] ,M[1][2] + M[3][4] +p[0]*p[2]*p[4] ,

M[1][3] + M[4][4] +p[0]*p[3]*p[4]}

{0+330+250 , 150+180+75 , 330+0+300} = 405

M[2][5] = min{M[2][2] + M[3][5] +p[1]*p[2]*p[5] ,M[2][3] + M[4][5] +p[1]*p[3]*p[5] ,

M[2][4] + M[5][5] +p[1]*p[4]*p[5]}

{0+930+1500 , 360+3000+6000,330+0+2500} = 2430

M[1][5] = min{M[1][1] +M[2][5]+p[0]*p[1]*p[5] , M[1][2] +M[3][5]+p[0]*p[2]*p[5],

M[1][3] +M[4][5]+p[0]*p[3]*p[5] , M[1][4] +M[5][5]+p[0]*p[4]*p[5]}

{0+2430+2500 , 150+930+750 , 330+3000+3000 , 405+0+1250} = 1655

(a)

MemoizedCutRod(p, n)

r: array(0..n) := (0 => 0, others =>MinInt)

return MemoizedCutRodAux(p, n, r)

MemoizedCutRodAux(p, n, r)

if r(n) = 0 and then n /= 0 then -- check if need to calculate a new solution

q: int := MinInt

for i in 1 .. n loop

q := max(q, p(i) + MemoizedCutRodAux(p, n-i, r))

end loop

end if

r(n) := q

end if

return r(n)

8 0

3 years ago

Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia

Mrrafil [7]

Answer:

The fin temperature in °C at a distance of 10 cm from the base = 33.78°C

Explanation:

The following assumptions will be made to solve this problem

- The heat transfer coefficient does not change with the time or distance.

- The temperature of the fins varies just in only one direction.

The temperature of the fin at x = 10 cm = 0.10 m from the base can be calculated from the temperature variation with distance formula for a very long fin.

(T - T∞) = (T₀ - T∞)e⁻ᵐˣ

T = T(x) = temperature at any point along the fin

T∞ = temperature at the tip of the fin = ambient temperature = 25°C

T₀ = temperature at the base of thw fin = 50°C

x = any distance along the length of the fin from the base of the fin = 0.1 m

m = √(hP/KA)

h = Heat transfer coefficient = 123 W/m².K

P = perimeter in contact with the base = πD = π × 0.03 = 0.0943 m

K = thermal conductivity = 150 W/m.K

A = surface area in contact with the base = πD²/4 = π(0.03)²/4 = 0.0007071 m²

m = √(123 × 0.0943)/(150 × 0.0007071)

m = 10.46

mx = 10.46 × 0.1 = 1.046

(T - 25) = (50 - 25) e⁻¹•⁰⁴⁶

T = 25 + 25 e⁻¹•⁰⁴⁶ = 25 + 8.78 = 33.78°C

8 0

3 years ago

Which contemporary jazz artist was one of the first to use a synthesizer in their recording

ivolga24 [154]

Answer:

In this era, Sun Ra was among the first of any musicians to make extensive and pioneering use of synthesizers and other various electronic keyboards; he was given a prototype Minimoog by its inventor, Robert Moog.

Explanation:

3 0

2 years ago