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9966 [12]
3 years ago
14

A four-lane freeway (two lanes in each direction) is located on rolling terrain and has 12-ft lanes, no lateral obstructions wit

hin 6 ft of the pavement edges, and there are 2 ramps within 3 miles upstream of the segment midpoint and 3 ramps within 3 miles downstream of the segment midpoint. A weekday directional peak-hour volume of 1800 vehicles (familiar users) is observed, with 700 arriving in the most congested 15-min period. If a level of service no worse than C is desired, determine the maximum number of heavy vehicles that can be present in the peak-hour traffic stream.
Consider the freeway and traffic conditions described in Problem 6.3. If 180 of the 1800 vehicles observed in the peak hour were heavy vehicles (assume a 70%/30% SUT/TT split), what would the level of service of this freeway be on a 5-mi, 6% downgrade?

Engineering
1 answer:
guapka [62]3 years ago
7 0

Answer:

Check the explanation

Explanation:

FFS = Free-Flow Speed (km/h) BFFS = the Base Free-Flow Speed (km/h) fLS = the Adjustment for lane and shoulder widths less than 3.65 m and 1.80 m, respectively. fAPD = Adjustment for access points density (km/h) fm = Adjustment for proportion  (km/h) 3.0 Results

Kindly check the below image for the step by step explanation to the above question.

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A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminu
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Answer:

a. 8 sheets of paper is needed between her plates to get the proper capacitance

b. Area of Aluminum Foil needed = 0.45m²

c. To keep a 1.0-nF, a larger area of Teflon is required.

Explanation:

a.

First, we need to calculate the distance between two plates.

This is given by

d = Kε0A/C

Where

K = 3

ε0 = Physical Constant = 8.854 * 10^-12 C²/Nm²

A = Area = 22 * 28 = 616cm² = 0.0616m²

C = 1.0-nF = 1 * 10^-12F

So, d = (3 * 8.854 * 10^-12 C²/Nm² * 0.0616) / (1 * 10^-12F)

d = 1.64 * 10^-3m

d = 1.64mm

Now, that the distance has been solved.

The Number of Sheets, N is given by

N = d/d,sheet where d, sheet =the sheet thickness = 0.2mm

N = 1.64/0.2

N = 8.2

N = 8 sheets --- Approximated

b.

Here, she's changed the diameter of the sheets to 12mm

Well make use of the formula in (a) above

Using d = Kε0A/C

Where

d = 12 * 10^-3m

Other constraints remain unchanged

Make A the subject of formula

A = dC/Kε0

A = (12 * 10^-3m * 1 * 10^-12F)/(3 * 8.854 * 10^-12 C²/Nm²)

A= 0.45m²

c. From (b) above

A ∝ 1/K

As the dielectric constant increase, the area decreases

The dielectric constant of a Teflon is 2.1

This means that if she used a Teflon instead, the area will be larger.

So, to keep a 1.0-nF, a larger area of Teflon is required.

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AveGali [126]

The current IDS is greater than 0 since the VGS has induced an inversion layer and the transistor is operating in the saturation region.

<u>Explanation:</u>

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  • A specific source voltage and gate of NMOS, the voltage get drained during the specific level, the drain voltage is rises beyond where there is no effect of current during saturated region.
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