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BabaBlast [244]
2 years ago
8

A horizontal curve of a two-lane undivided highway (12-foot lanes) has a radius of 678 feet to the center line of the roadway. A

n old building (sight obstruction) is located 30 feet from the edge of the innermost lane. The road is level and the superelevation is 0.06. Please determine the maximum speed for safe vehicle operation on this horizontal curve.
Engineering
1 answer:
OLEGan [10]2 years ago
5 0

Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

radius ( R ) = 678 ft

old building located ( m )= 30 ft

super elevation = 0.06

<u>Determine the maximum speed for safe vehicle operation </u>

firstly calculate the stopping sight distance

m = R ( 1 - cos \frac{28.655*S}{R} )  ----  ( 1 )

R = 678  

m ( horizontal sightline ) = 30 ft

back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) )  = 30 / 678 = 0.044

cos \frac{28.65 *s }{678}  = 1.044

hence ; 28.65 * s = 678 * 0.2956

s = 6.99 ≈ 7 ft

next we will calculate the design speed ( u ) using the formula below

S = 1.47 ut  + \frac{u^2}{30(\frac{a}{3.2} )-G1}  ----  ( 2 )

t = reaction time,  a = vehicle acceleration, G1 = grade percentage

assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0

back to equation 2

6.99 = 1.47 * u * 2.5 + \frac{u^2}{30[(11.2/32.2)-0 ]}

3.675 u  + 0.0958 u^2 - 6.99 = 0

u ( 3.675 + 0.0958 u ) = 6.99

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Veseljchak [2.6K]

Answer:

Hello your question is incomplete attached below is the missing part and answer

options :

Effect A

Effect B

Effect C

Effect D

Effect AB

Effect AC

Effect AD

Effect BC

Effect BD

Effect CD

Answer :

A  = significant

 B  = significant

C  = Non-significant

D  = Non-significant

AB  = Non-significant

AC  = significant

AD  = Non-significant

BC  = Non-significant

 BD  = Non-significant

 CD = Non-significant

Explanation:

The dependent variable here is Time

Effect of A  = significant

Effect of B  = significant

Effect of C  = Non-significant

Effect of D  = Non-significant

Effect of AB  = Non-significant

Effect of AC  = significant

Effect of AD  = Non-significant

Effect of BC  = Non-significant

Effect of BD  = Non-significant

Effect of CD = Non-significant

8 0
3 years ago
Which of the following requires formwork?
vekshin1

Answer:

D

Explanation:

Masonry uses stone work, making a stone wall requires perfect masonry.

3 0
2 years ago
Consider a Carnot heat pump cycle executed in a steady-flow system in the saturated mixture region using R-134a flowing at a rat
attashe74 [19]

Answer:

7.15

Explanation:

Firstly, the COP of such heat pump must be measured that is,

              COP_{HP}=\frac{T_H}{T_H-T_L}

Therefore, the temperature relationship, T_H=1.15\;T_L

Then, we should apply the values in the COP.

                           =\frac{1.15\;T_L}{1.15-1}

                           =7.67

The number of heat rejected by the heat pump must then be calculated.

                   Q_H=COP_{HP}\times W_{nst}

                          =7.67\times5=38.35

We must then calculate the refrigerant mass flow rate.

                   m=0.264\;kg/s

                   q_H=\frac{Q_H}{m}

                         =\frac{38.35}{0.264}=145.27

The h_g value is 145.27 and therefore the hot reservoir temperature is 64° C.

The pressure at 64 ° C is thus 1849.36 kPa by interpolation.

And, the lowest reservoir temperature must be calculated.

                   T_L=\frac{T_H}{1.15}

                        =\frac{64+273}{1.15}=293.04

                        =19.89\°C

the lowest reservoir temperature = 258.703  kpa                    

So, the pressure ratio should be = 7.15

8 0
3 years ago
A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric c
Paraphin [41]

Answer:

10.8\ \text{lb/ft^2}

101.96\ \text{lb/ft}^2

Explanation:

v_1 = Velocity of car = 65 mph = 65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}

\rho = Density of air = 0.00237\ \text{slug/ft}^3

v_2=0

P_1=0

h_1=h_2

From Bernoulli's law we have

P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}

The maximum pressure on the girl's hand is 10.8\ \text{lb/ft^2}

Now v_1 = 200 mph = 200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}

P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2

The maximum pressure on the girl's hand is 101.96\ \text{lb/ft}^2

5 0
3 years ago
A fisherman tries to escape a flash flood by walking up the 10 degree hill. If the water is rising at 0.11 m/s, how fast must he
FromTheMoon [43]

10.5 mph I’m pretty sure so so sorry if I’m wrong

6 0
3 years ago
Read 2 more answers
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