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aleksandr82 [10.1K]
3 years ago
11

A six-lane multilane highway (three lanes in each direction) has a peak-hour factor of 0.90, 11-ft lanes with a 4-ft right-side

shoulder, and a two-way left-turn lane in the median. The directional peak-hour traffic flow is 4000 vehicles with 14% SUTs and 6% TTs. What will the level of service of this highway be on a 4% upgrade that is 1.5 miles long if the speed limit is 55 mi/h and there are 15 access points per mile

Engineering
1 answer:
Dennis_Churaev [7]3 years ago
4 0

Answer:

40.04pc/mile/hour

Explanation:

Check the attachment

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A manufacturer makes two types of drinking straws: one with a square cross-sectional shape, and the other type the typical round
Harlamova29_29 [7]

Answer:

\frac{Q_{square}}{Q_{circle}} =  0.785  

Explanation:

given data

types of drinking straws

  1. square cross-sectional shape
  2. round shape

solution

we know that both perimeter of the cross section are equal

so we can say that

perimeter of square  = perimeter of circle  

4 × S = π × D

here S is length and D is diameter

S = \frac{\pi D}{4}        ....................1

and

ratio of  flow rate through the square and circle is here

\frac{Q_{square}}{Q_{circle}} = \frac{AV^2}{AV^2}  

\frac{Q_{square}}{Q_{circle}} = \frac{S^2}{\frac{\pi D^2}{4}}  

\frac{Q_{square}}{Q_{circle}} = \frac{(\frac{\pi D}{4})^2}{\frac{\pi D^2}{4}}  

\frac{Q_{square}}{Q_{circle}} = \frac{\pi }{4}  

\frac{Q_{square}}{Q_{circle}} =  0.785  

4 0
4 years ago
If engineering is easy then why don't most people join?
Mazyrski [523]
Engineering requires a lot of school
5 0
3 years ago
Read 2 more answers
Calculate the wire pressure for a round copper bar with an original cross-sectional area of 12.56 mm2 to a 30% reduction of area
dybincka [34]

Answer:153.76 MPa

Explanation:

Initial Area\left ( A_0\right )=12.56 mm^2

Final Area\left ( A_f\right )=0.7\times 12.56 mm^2=8.792 mm^2

Die angle=30^{\circ}

\alpha =\frac{30}{2}=15^{\circ}

\mu =0.08

Yield stress\left ( \sigma _y \right )=350 MPa

B=\mu cot\left ( \aplha\right )=0.2985

\sigma _{pressure}=\sigma _y\left [\frac{1+B}{B}\right ]\left [ 1-\frac{A_f}{A_0}\right ]^B

\sigma _{pressure}=350\left [\frac{1+0.2985}{0.2985}\right ]\left [ 1-\frac{8.792}{12.56}\right ]^{0.2985}

\sigma _{pressure}=153.76 MPa

8 0
3 years ago
Why does the compression-refrigeration cycle have a high-pressure side and a low-pressure side?
Cloud [144]

Answer: D

Explanation:

8 0
2 years ago
The output side of an ideal transformer has 35 turns, and supplies 2.0 A to a 24-W device. Ifthe input is a standard wall outlet
Crank

Answer:

The current drawn from the outlet is 0.2 A

The number of turns on the input side is 350 turns

Explanation:

Given;

number of turns of the secondary coil, Ns = 35 turns

the output current, I_s = 2 A

power supplied, P_s = 24 W

the standard wall outlet in most homes = 120 V = input voltage

For an ideal transformer; output power = input power

the current drawn from the outlet is calculated;

I_pV_p = P_s\\\\I_p = \frac{P_s}{V_p} = \frac{24}{120} = 0.2 \ A

The number of turns on the input side is calculated as;

\frac{N_p}{N_s} = \frac{I_s}{I_p}  \\\\N_p = \frac{N_sI_s}{I_p} \\\\N_p = \frac{35 \times 2}{0.2} \\\\N_p = 350 \ turns

4 0
3 years ago
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