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storchak [24]
4 years ago
11

When two parallel and coplanar shafts are connected by gears having teeth parallel to the axis of the shaft, the arrangement is

known as spiral gearing.
a) true
b) false
Engineering
1 answer:
Gelneren [198K]4 years ago
8 0

Answer:

False

Explanation:

The spiral gearing is the application that is commonly used in any vehicle where the drive from the shaft will turn perpendicular to the drive of the wheel.

The two parallel and coplanar shafts are connected by gears having helical teeth perpendicular to the axis of the shaft.

Spiral gearing application is appropriate for any machine or vehicle with a demand of great velocity and large torque power.

When two parallel and coplanar shafts are connected by gears having teeth parallel to the axis of the shaft, the arrangement is known as spiral gearing is therefore false.

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A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula
makkiz [27]

Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

We are given;

Length; L = 5ft = 1.52.

Initial diameter;d1 = 12in = 0.3m

Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

6 0
3 years ago
Suppose we are managing a consulting team of expert computer hackers, and each week we have to choose a job for them to undertak
lina2011 [118]

Answer:

if number == 1

  then

  tempSolution= max(l[number],h[number])

else if number == 2 then

  tempSolution= max(optimalPlan(1, l, h)+ l[2], h[2])

else

  tempSolution= max(optimalPlan(number − 1, l, h) + l[number], optimalPlan(number − 2, l, h) + h[number])

end if

return Value

FindOptimalValue(number, l, h)

for itterator = 1 ! number do

  tempSolution[itterator] = 0

end for

for itterator = 1 ! number do

  if itterator == 1 then

      tempSolution[itterator] max(l[itterator], h[itterator])

  else if itterator == 2 then

      tempSolution[itterator] max(tempSolution[1] + l[2], h[2])

  else

      tempSolution[itterator] max(tempSolution[itterator − 1] + l[itterator], tempSolution[itterator − 2] + h[itterator])

  end if

end for

return Value[number]

OPtimalPlan(number, l, h, Value)

for itterator = 1 ! number do

  WeekVal[itterator]

end for

if tempSolution[number] − l[number] = tempSolution[number − 1] then

  WeekVal[number] ”Low stress”

  OPtimalPlan(number-1, l, h, Value)

else

  WeekVal[number] ”High stress”

  OPtimalPlan(number-2, l, h, Value)

end if

return WeekVal

7 0
3 years ago
Estimate the energy (head) loss a short length of a pipe conveying 300 litres of water per second and suddenly enlarging from a
Ludmilka [50]

Known :

Q = 300 L/s = 0.3 m³/s

D1 = 350 mm = 0.35 m

D2 = 700 mm = 0.7 m

g = 9.81 m/s²

Solution :

A1 = πD1² / 4 = π(0.35²) / 4 = 0.096 m²

A2 = πD2² / 4 = π(0.7²) / 4 = 0.385 m²

hL = (kL / 2g) • (U1² - U2²)

hL = (kL / 2g) • Q² (1/A1² - 1/A2²)

hL = (1 / 2(9.81)) • (0.3²) • (1/(0.096²) - 1/(0.385²))

hL = 0.467 m

5 0
3 years ago
How is a Doctor Who is a generalist different a specialist?
bekas [8.4K]

Answer:

specialist focus in one very specific thing, generalist focus and many things, they are like a jack of all trades

7 0
3 years ago
Poles are values of Laplace transform variable, s, that make denominator of transfer function zero. Zeros are values of Laplace
Ostrovityanka [42]

Answer:

Zero 1 = -1

Zero 2 = -3

Pole 1 = 0

Pole 2 = -2

Pole 3 = -4

Pole 4 = -6

Gain = 4

Explanation:

For any given transfer function, the general form is given as

T.F = k [N(s)] ÷ [D(s)]

where k = gain of the transfer function

N(s) is the numerator polynomial of the transfer function whose roots are the zeros of the transfer function.

D(s) is the denominator polynomial of the transfer function whose roots are the poles of the transfer function.

k [N(s)] = 4s² + 16s + 12 = 4[s² + 4s + 3]

it is evident that

Gain = k = 4

N(s) = (s² + 4s + 3) = (s² + s + 3s + 3)

= s(s + 1) + 3 (s + 1) = (s + 1)(s + 3)

The zeros are -1 and -3

D(s) = s⁴ + 12s³ + 44s² + 48s

= s(s³ + 12s² + 44s + 48)

= s(s + 2)(s + 4)(s + 6)

The roots are then, 0, -2, -4 and -6.

Hope this Helps!!!

3 0
4 years ago
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