1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
iragen [17]
3 years ago
13

A minor road intersects a major 4-lane divided road with a design speed of 50 mph and a median width of 12 ft. The intersection

is controlled with a stop sign on the minor road. If the design vehicle is a passenger car, determine the minimum sight distance required on the major road that will allow a stopped vehicle on the minor road to safely turn left if the approach grade on the minor road is 5%.
Engineering
1 answer:
satela [25.4K]3 years ago
8 0

Answer:

minimum sight distance = 699 ft

Explanation:

given data

road lane = 4 divided road

median width = 12 ft

grade road = 5%

solution

we take here time gap factor for minor road vehicle when enter to major road  from table

time gap = 8.1  sec

and for median width of 12 ft

time gap = 8.2 + 0.7 ( 1 + \frac{12}{12}  )  

time gap = 9.5 second

so minimum sight distance will be

minimum sight distance = 1.47 × design speed × time gap  

minimum sight distance = 1.47 × 50  × 9.5

minimum sight distance = 699 ft

You might be interested in
A heat pump receives heat from a lake that has an average wintertime temperature of 6o C and supplies heat into a house having a
Dafna1 [17]

Answer:

a) \dot W = 1.062\,kW

Explanation:

a) Let consider that heat pump is reversible, so that the Coefficient of Performance is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

COP_{HP} = \frac{298.15\,K}{298.15\,K-279.15\,K}

COP_{HP} = 15.692

The minimum heat received by the house must be equal to the heat lost to keep the average temperature constant. Hence:

\dot Q_{H} = 60000\,\frac{kJ}{h}

The minimum power supplied to the heat pump is:

\dot W = \frac{\dot Q_{H}}{COP}

\dot W = \frac{\left(60000\,\frac{kJ}{h}  \right)\cdot \left(\frac{1\,h}{3600\,s}  \right)}{15.692}

\dot W = 1.062\,kW

5 0
3 years ago
Can a 1½ " conduit, with a total area of 2.04 square inches, be filled with wires that total 0.93 square inches if the maximum f
Papessa [141]

Answer:

it is not possible to place the wires in the condui

Explanation:

given data

total area = 2.04 square inches

wires total area = 0.93 square inches

maximum fill conduit =  40%

to find out

Can it is possible place wire in conduit conduit

solution

we know maximum fill is 40%

so here first we get total area of conduit that will be

total area of conduit = 40% × 2.04

total area of conduit = 0.816 square inches

but this area is less than required area of wire that is 0.93 square inches

so we can say it is not possible to place the wires in the conduit

4 0
3 years ago
Can you solve this question​
Alecsey [184]

Answer:

eojcjksjsososisjsiisisiiaodbjspbcpjsphcpjajosjjs ahahhahahahahahahahahahahahahhhahahahaahahhahahahahaahahahahaha

6 0
2 years ago
Read 2 more answers
What is the term RF exiciter?
Sati [7]
The exciter provides fully coherent receiver local oscillator signals at radar frequency band as well as requisite, auxiliary high frequency clock signals. The exciter function is divided into an internal frequency synthesizer and an upconverter.

Hope this helps :)))
7 0
2 years ago
10% A steel beam W18x76 spans 32 feet and is subjected to a Moment of 334 kips-ft. Find the load w on the beam. Determine the de
Lilit [14]

Answer:

w = 10.437 kips

deflection at 1/4 span  20.83\E ft

at mid span = 1.23\E ft

shear stress  7.3629 psi

Explanation:

area of cross section = 18*76

length of span = 32 ft

moment = 334 kips-ft

we know that

moment = load *eccentricity

334 = w * 32

w = 10.437 kips

deflection at 1/4 span

\delta = \frac{wa^2b^2}{3EI}

= \frac{10.4375*8^2 *24^2}{3E \frac{BD^3}{12}}

         =\frac{10.437 *8^2*24^2}{3E \frac{18*16^3}{12}}

         = 20.83\E ft

at mid span

\delta = \frac{wl^3}{48EI}

= \frac{10.43 *32^3}{48 *E*\frac{18*16^3}{12}}

\delta = 1.23\E ft

shear stress

\tau = \frac{w}{A} = \frac{10.43 7*10^3}{18*76} =7.3629 psi

6 0
3 years ago
Other questions:
  • How do Solar Engineers Help Humans?<br> (2 or more sentences please)
    9·1 answer
  • What are the desired characteristics or values for the following parameters of an ideal amplifier?o Phase change as a function o
    10·1 answer
  • What are common names assigned to instruction addresses in a PLC program called
    8·1 answer
  • What is the most likely reason the rover won't travel in a straight line?
    9·1 answer
  • What was the most important thing you learned this school year in your engineering class and why did you choose this thing
    15·1 answer
  • Just need someone to talk to pls dont just use me for points
    5·1 answer
  • Two children are playing on a seesaw. The child on the left weighs 50 lbs. And the child on the right weighs 100 lbs. If the chi
    5·1 answer
  • Reverse Engineering: Structural Analysis
    14·1 answer
  • A high compression ratio may result in;
    13·1 answer
  • Which of the following situations best describes student engaged in active learning
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!