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katrin2010 [14]
3 years ago
5

A multilane highway (two lanes in each direction) is on level terrain. The free-flow speed has been measured at 45 mi/h. The pea

k-hour directional traffic flow is 1300 vehicles with 8% heavy vehicles. If the peak-hour factor is 0.85, determine the highway's level of service.
Engineering
1 answer:
Pani-rosa [81]3 years ago
7 0

Answer:

17.71 veh/mi/ln

Explanation:

Given,

Free Flow Speed, FFS = 45 mph

Number of lanes in each direction = 2

Peak Flow, V = 1300 veh/hr

Peak-hour factor = 0.85

HV = 8 %

Level Terrain

fHV = 1/ (1 + 0.08 (1.5-1)) = 1/1.04 = 0.96

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 1300/(0.96*2*0.85*1.0) = 796.57 ~ 797 veh/hr/ln

Vp < 1400

S = FFS = 45 mph

Density = Vp/S = (797) / (45) = 17.71 veh/mi/ln

Density of LOS B should lie between 11 – 18 veh/mi/ln

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What is heat unit?in X ray machine​
Karo-lina-s [1.5K]

Answer:

joule

Explanation:

heat is express in joules in x Ray equipment

6 0
3 years ago
: A freeway exit ramp has a single lane and consist of entirely of a horizontal curve with a central angle of 90 degree and a le
Mariana [72]

The design speed was used for the freeway exit ramp is 11 mph.

<h3>Design speed used in the exit ramp</h3>

The design speed used in the exit ramp is calculated as follows;

f = v²/15R - 0.01e

where;

  • v is designated speed

v = ωr

v = (θ/t) r

θ = 90⁰ = 1.57 rad

v = (1.57 x 19.4)/2.5 s

v = 12.18 ft/s = 8.3 mph

<h3>Design speed</h3>

f = v²/15R - 0.01e

let the maximum superelevation, e = 1%

f = (8.3)²/(15 x 19.4) - 0.01

f = 0.22

0.22 is less than value of f which is 0.4

<h3>next iteration, try 10 mph</h3>

f = (10)²/(15 x 19.4) - 0.01

f = 0.33

0.33 is less than 0.4

<h3>next iteration, try 11 mph</h3>

f = (11)²/(15 x 19.4) - 0.01

f = 0.4

Thus, the design speed was used for the freeway exit ramp is 11 mph.

Learn more about design speed here: brainly.com/question/22279858

#SPJ1

4 0
2 years ago
The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta
tatyana61 [14]

Answer:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular  momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

H_o =r x mv=rxL

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change  of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular  momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2".

If we analyze the staritning point we see that the initial velocity can be founded like this:

v_o =\omega r_{OIC}=\omega (0.15m)

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af

0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)

And if we integrate the left part and we simplify the right part we have

1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega

And if we solve for \omega we got:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

8 0
3 years ago
Please read the short case at the end of Chapter 8, "Mary Barra of General Motors Values Culture". Use the following prompts to
Tom [10]

Answer: Incoherent question

Explanation: This is an act of plagiarism at subjecting the tutor to unnecessary stress at answering the purported question.

4 0
3 years ago
Which of the following Python lines calculates the standard deviation of variables Exam1 and Exam4 from a CSV file called ExamSc
Virty [35]

Answer:

The option B calculates the standard deviation of variables Exam1 and Exam4.

Explanation:

First, let's write the first two lines correctly:

  1. import pandas as pd
  2. scores = pd.read_csv('ExamScores.csv')

Now, let's analyse each option:

- The option A is incorrect

The expression std[['Exam1','Exam4']] is the incorrect part because in this case std is working as a variable and the std was not previously declared. So it will generate an error.

- The option C is incorrect

The expression std[['Exam1','Exam4']].scores() is incorrect because of two reasons: first, std is working as a variable and the std was not previously declared (as in option A), second, .scores() is working as a method and in this case, scores is the variable. So it will generate an error.

- The option D is incorrect

The expression [['Exam1']] is the incorrect part because it doesn't include the 'Exam4' and the problem is asking for the standard deviation of variables Exam1 and Exam4.

- The option B is CORRECT

The expression std[['Exam1','Exam4']] refers to the variables Exam1 and Exam4 from the total exams. As scores is a pandas variable and .std() is a method provided by pandas library, the expression .std() allows to get the deviation standard of the variables Exam1 and Exam4.

Thus, the option B calculates the standard deviation of variables Exam1 and Exam4.

4 0
3 years ago
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