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steposvetlana [31]
3 years ago
9

A gearbox is needed to provide an exact 30:1 increase in speed, while minimizing the

Engineering
1 answer:
alekssr [168]3 years ago
4 0

Answer:

answer

Explanation:

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Assume we have one road section which has 3 lanes in both directions. If the Sf for both direction is 75 mph, and Dj for both di
givi [52]

Answer:

i) 3750 veh/hr/ln

ii) 100 veh/mi/In

iii) 37.5 mph

Explanation:

number of lanes = 3

sf for both directions = 75 mph  ( free mean speed )

Dj for both directions = 200 veh/mi/In

<u>Calculate the value of  S0, D0 (veh/mi/ln) and maximum Vm (veh/hr)</u>

For either direction we will consider the total volume = 3 lanes

value of Dj = 3 lanes * 200 = 600 veh/mi/

i) value of SO

=  ( Dj * sf ) / 4 = ( 600 * 75 ) / 4 = 11250 veh/hr  = 3750 veh/hr/lane

ii) Value of DO

DO = Dj / 2 = 200 /2 = 100 veh/mi/In

iii) Value of Vm

= sf /2 = 75 / 2 = 37.5 mph

5 0
3 years ago
The condensed Q formula may be used for operations in which the friction loss can be determined for:
yan [13]

The condensed Q formula may be used for operations in which the friction loss can be determined for a: 3, 4, or 5 inch hose.

<h3>What is a firehose friction loss?</h3>

A firehose friction loss can be defined as a measure of the effect of the resistance of water against the inner side of a firehose, which typically results in a pressure drop at the terminal end.

Generally, some of the factors that affect the resistance or friction in a firehose include:

  • Length of hose.
  • Age of hose.
  • Water flow (gpm)
  • Water turbulence
  • Gravity

Mathematically, the firehose friction loss can be calculated by using this formula:

FL = C × (Q/100)² × L/100.

Read more on friction loss here: brainly.com/question/17305262

#SPJ1

3 0
2 years ago
What two units of measurement are used to classify engine sizes?
seraphim [82]
Liters or cubic inches
8 0
3 years ago
Read 2 more answers
Best budget freestyle drone?
RUDIKE [14]

Hx115 it's pretty fine looking

4 0
3 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in
RideAnS [48]

To resolve this problem we have,

R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm

F_{f2} is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa

After that, we can calculate the flexural strenght for the square cross section using the previously value.

\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN

6 0
4 years ago
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