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Mama L [17]
4 years ago
8

A segment of four-lane freeway (two lanes in each direction) has a 3% upgrade that is 1500 ft long followed by a 1000-ft 4% upgr

ade. It has 12-ft lanes and 3-ft shoulders. The directional hourly traffic flow is 2000 vehicles with 3% SUTs and 3% TTs. The total ramp density for this freeway segment is 2.33 ramps per mile. If the peak-hour factor is 0.90, what is the level of service of this compound-grade freeway segment

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

Answer:

The level of service of of compound grade freeway is LOSB.

Explanation:

Find the provided attachments for explanation

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sashaice [31]

The reason why giant stars become planetary nebulas is  Supergiant stars do not have enough mass to generate the gravity necessary to cause a planetary nebula.

<h3>Why do giant stars become planetary nebulae?</h3>

A planetary nebula is known to be formed or created by a dying star. A red giant is known to be unstable and thus emit pulses of gas that is said to form a sphere around the dying star and thus they are said to  be ionized by the ultraviolet radiation that the star is known to releases.

Learn more about  giant stars from

brainly.com/question/27111741

#SPJ1

3 0
2 years ago
A liquid refrigerant (sg=1,2) is flowing at a weight flow rate of 20,9 N/h. Refrigerant flashes into a vapor and its specific we
Iteru [2.4K]

Answer:

Explanation:

volume of 20.9 N

= 20.9 / 11.5 m³

= 1.8174 m³

In one hour 1.8174 m³ flows

in one second volume flowing = 1.8174 / 60 x 60

= 5 x 10⁻⁴ m³

Rate of volume flow = 5 x 10⁻⁴ m³ / s .

5 0
3 years ago
Create two arrays with 5 elements each: one will hold Strings and the second will hold integers. Write a program to ask the user
MrMuchimi

Answer:

#include <iostream>

#include <iomanip>

#include <string>

using namespace std;

int main() {

   string name[5];  

   int age[5];  

   int i,j;  

   for ( i = 0; i<=4; i++ ) {  

       cout << "Please enter student's name:";  

       cin >> name[i];  

       cout << "Please enter student's age:";  

       cin >> age[i];          

   }  

for (i=0;i<=4;i++){

   cout<<"Age of  "<< name[i]<<"  is  "<<age[i]<<endl;  

}

}

Output of above program is displayed in figure attached.

5 0
3 years ago
The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in
Mazyrski [523]

Explanation:

A.

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

B.

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

5 0
3 years ago
What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin
DochEvi [55]

Answer:

m_{LP}=0.45\,kg

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]

Q_{water} = 3599.435\,kJ

The heat liberated by the LP gas is:

Q_{LP} = \frac{3599.435\,kJ}{0.16}

Q_{LP} = 22496.469\,kJ

A kilogram of LP gas has a minimum combustion power of 50028\,kJ. Then, the required mass is:  

m_{LP} = \frac{22496.469\,kJ}{50028\,\frac{kJ}{kg} }

m_{LP}=0.45\,kg

6 0
3 years ago
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