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LenaWriter [7]
3 years ago
12

A divided multilane highway in a recreational area has four lanes (two lanes in each direction) and is on rolling terrain. The h

ighway has 10-ft lanes with a 5-ft right-side shoulder and a 3-ft left-side shoulder. The posted speed is 55 mi/h. Previously, there were six access points per mile, but recent development has increased the number of access points to 12 per mile. Before development, the peak-hour factor was 0.95, and the directional hourly volume was 2300 vehicles with 15% heavy vehicles. After development, the peak-hour directional flow is 2700 vehicles with the same vehicle percentages and peak-hour factor.
1. What is the LOS before and after the development?
Engineering
1 answer:
sertanlavr [38]3 years ago
8 0

Answer:

Explanation:

Before: PT= 0.10, PB= 0.03 (given) ET = 2.5 ER = 2.0 (Table 6.5)

fHV= 0.847 (Eq. 6.5) PHF = 0.95, fp= 0.90, N=2, V = 2200 (given)

vp= V/[PHF⋅fHVTB⋅fp⋅N] = 1518.9 (Eq. 6.3)

BFFS = 50+5, BFFS =55 (given) fLW= 6.6

TLC=6+3=9 fLC= 0.65

fM= 0.0

fA= 1.0

FFS = BFFS −fLW−fLC–fM–fA= 46.75 (Eq. 6.7)

Use FFS=45 D= vp/S = 33.75pc/mi/ln Eq (6.6)

After: fA= 3.0

FFS = BFFS −fLW−fLC–fM–fA= 44.75 (Eq. 6.7)

Use FFS=45 Vnew= 2600 Vp= Va/[PHF⋅fHB⋅fp⋅N] = 1795 (Eq. 6.3) D= vp/S = 39.89pc/mi/ln

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Electric current originates from which part of an atom? *
yanalaym [24]

Answer: Electric current originates from positively charged protons negatively charged electrons of an atom.

Explanation:

The movement of ions (positive or negative) from one point to another is called electric current.

An atom has three sub-atomic particles. These are protons, neutrons and electrons.

Protons are positively charged, neutrons have no charge and electrons are negatively charged. Protons and neutrons reside inside the nucleus of an atom whereas electrons revolve around the nucleus.

So, protons and electrons are responsible for originating electric current form an atom as these are the charged particles.

Thus, we can conclude that electric current originates from positively charged protons negatively charged electrons of an atom.

3 0
3 years ago
What is the purpose of gears?
Brrunno [24]

Answer:

To help wheels move in a circle

Explanation:

6 0
3 years ago
Determine the period of each of the following discrete-time signals (if a signal is not periodic, denote its period by infinity)
sergiy2304 [10]

Answer:

a) it is periodic

N = (20/3)k = 20 { for K =3}

b) it is Non-Periodic.

N = ∞

c) x(n) is periodic

N = LCM ( 5, 20 )

Explanation:

We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.

then the period of the signal is given as

N = ( 2π/w₀)K

k is least integer for which N is also integer

Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2

now

a) cos(2π(0.15)n)

w₀ = 2π(0.15)

Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3

so, it is periodic

N = (20/3)k = 20 { for K =3}

b) cos(2n);

w₀ = 2

Now, 2π/w₀ = 2π/2) = π

so, it is Non-Periodic.

N = ∞

c)  cos(π0.3n) + cos(π0.4n)

x(n) = x1(n) + x2(n)

x1(n) = cos(π0.3n)

x2(n) =  cos(π0.4n)

so

w₀ = π0.3

2π/w₀  = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3

∴ N1 = 20

AND

w₀ = π0.4

2π/w₀  = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5

∴ N² = 5

so, x(n) is periodic

N = LCM ( 5, 20 )

6 0
3 years ago
The following are the results of a sieve analysis. U.S. sieve no. Mass of soil retained (g) 4 0 10 18.5 20 53.2 40 90.5 60 81.8
il63 [147K]

Answer:

a.)

US Sieve no.                         % finer (C₅ )

4                                                  100

10                                                95.61

20                                               82.98

40                                               61.50

60                                               42.08

100                                              20.19

200                                              6.3

Pan                                               0

b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4

c.) Cu = 3.33

d.) Cc = 1

Explanation:

As given ,

US Sieve no.             Mass of soil retained (C₂ )

4                                            0

10                                          18.5

20                                         53.2

40                                         90.5

60                                         81.8

100                                        92.2

200                                       58.5

Pan                                        26.5

Now,

Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g

⇒ w = 421.2 g

As we know that ,

% Retained = C₃ = C₂×\frac{100}{w}

∴ we get

US Sieve no.               % retained (C₃ )               Cummulative % retained (C₄)

4                                            0                                           0

10                                          4.39                                      4.39

20                                         12.63                                     17.02

40                                         21.48                                     38.50

60                                         19.42                                     57.92

100                                        21.89                                     79.81

200                                       13.89                                     93.70

Pan                                        6.30                                      100

Now,

% finer = C₅ = 100 - C₄

∴ we get

US Sieve no.               Cummulative % retained (C₄)          % finer (C₅ )

4                                                     0                                          100

10                                                  4.39                                      95.61

20                                                 17.02                                     82.98

40                                                 38.50                                    61.50

60                                                 57.92                                    42.08

100                                                79.81                                     20.19

200                                                93.70                                   6.3

Pan                                                 100                                        0

The grain-size distribution is :

b.)

From the diagram , we can see that

D10 = 0.12

D30 = 0.22

D60 = 0.12

c.)

Uniformity Coefficient = Cu = \frac{D60}{D10}

⇒ Cu = \frac{0.4}{0.12} = 3.33

d.)

Coefficient of Graduation = Cc = \frac{D30^{2}}{D10 . D60}

⇒ Cc = \frac{0.22^{2}}{(0.4) . (0.12)} = 1

3 0
3 years ago
What are the minimum and maximum required footing projections for a concrete footing having a thickness of 8 inches
SpyIntel [72]

The minimum and maximum required footing projections for a concrete footing having a thickness of 8 inches will be 4 inches and 8 inches.

<h3>What is projection?</h3>

It should be noted that footing projections are important for construction purposes

In this case, the minimum and maximum required footing projections for a concrete footing having a thickness of 8 inches will be 4 inches and 8 inches.

Learn more about projection on:

brainly.com/question/3703881

#SPJ12

8 0
2 years ago
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