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Y_Kistochka [10]
3 years ago
5

A developer has requested permission to build a large retail store at a location adjacent to the intersection of an undivided fo

ur-lane major road and a two-lane minor road. Traffic on the minor road is controlled by a stop sign. The speed limits are 35 and 50 mi/h on the minor and major roads, respectively. The building is to be located at a distance of 65 ft from the near lane of one of the approaches of the minor road. Determine the minimum sight distance required on the major road for the stopped vehicle on the minor road to safely turn right onto the major road. Design vehicle is a single-unit truck. Lanes on the major road are 12 ft wide. Hint: Adjust for the number of lanes. Given: Time gap for single-unit truck for Right turn = 8.5 sec Time adjustment for number of lanes = 0.7 sec
Engineering
1 answer:
Sergio [31]3 years ago
6 0

Answer:

676 ft

Explanation:

Minimum sight distance, d_min

d_min = 1.47 * v_max * t_total where v_max is maximum velocity in mi/h, t_total is total time

v_max is given as 50 mi/h

t_total is sum of time for right-turn and adjustment time=8.5+0.7=9.2 seconds

Substituting these figures we obtain d_min=1.47*50*9.2=676.2 ft

For practical purposes, this distance is taken as 676 ft

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Alex17521 [72]

Answer:

do it if it's Sunday why are you like this

6 0
3 years ago
At the inlet to the combustor of a supersonic combustion ramjet (or scramjet), the flow Mach number is supersonic. For a fuel-ai
Daniel [21]

Answer:

The inlet match number is 3.42.

Explanation:

8 0
3 years ago
A driver is traveling at 90 km/h down a 3% grade on good, wet pavement. An accident
Paul [167]

Answer:

0.35

Explanation:

We resolve the component of the weight of the car along and perpendicular to the grade. We have mgsinФ and mgcosФ where Ф = angle of grade.

Now, the frictional force f = μN = μmgcosФ where μ = coefficient of friction

So, the net force along the grade is F = mgsinФ - μmgcosФ.

The work done by this force moving a distance, d along the grade is

W = (mgsinФ - μmgcosФ)d

This work equals the change in kinetic energy of the car. So ΔK = 1/2m(v₂² - v₁²) = W = (mgsinФ - μmgcosФ)d

1/2m(v₂² - v₁²) = (mgsinФ - μmgcosФ)d

1/2(v₂² - v₁²) = (gsinФ - μgcosФ)d

(v₂² - v₁²)/2d = (gsinФ - μgcosФ)

dividing through by gcosФ, we have

(v₂² - v₁²)/2dgcosФ = (gsinФ/gcosФ) - μgcosФ/gcosФ

(v₂² - v₁²)/2dgcosФ = tanФ -  μ

μ = tanФ - (v₂² - v₁²)/2dgcosФ

given that tanФ = 3% = 3/100 and 1 + tan²Ф = 1/cos²Ф, cosФ = 1/(√1 + tan²Ф) = 1/(√1 + (3/100)²) = 1/(√1 + (9/10000)) = 1/(√10000 + 9/10000) = 1/√(10009/10000) = 100/√10009 = 100/100.05 = 0.9995.

Also, given that v₁ = 90 km/h = 90 × 1000/3600 m/s = 25 m/s and v₂ = 45 km/h = 45 × 1000/3600 m/s = 12.5 m/s, d = 75 m and g = 9.8 m/s².

So, substituting the values of the variables into the equation, we have

μ = tanФ - (v₂² - v₁²)/2dgcosФ

μ = 3/100 - ((12.5 m/s)² - (25 m/s)²)/(2 × 75 m × 9.8 m/s² × 0.9995)

μ = 3/100 - ((156.25 m/s)² - (625 m/s)²)/1,469.265 m²/s²

μ = 3/100 - (-468.75 m²/s²)/1,469.265 m²/s²

μ = 3/100 + 468.75 m²/s²/1,469.265 m²/s²

μ = 0.03 + 0.32

μ = 0.35

So, theoretical friction  coefficient is 0.35

4 0
3 years ago
List the parts of a manual transmission <br><br> List the parts of a typical clutch assembly?
True [87]

Answer:

Explanation: Clutch Plate.

Clutch Cover.

Clutch Bearing (Release bearing)

Release Fork (clutch fork)

7 0
2 years ago
As the impurity concentration in solid solution of a metal is increased, the tensile strength:________.a) decreasesb) increasesc
valkas [14]

Answer:

Increases

Explanation:

By inhibiting the motion of dislocations by impurities in a solid solutions, is a strengthening mechanism. In solid solutions it is atomic level strengthening resulting from resistance to dislocation motion. Hence, the strength of the alloys can differ with respect to the precipitate's property. Example, the precipitate is stronger (ability to an obstacle to the dislocation motion) than the matrix and it shows an improvement of strength.

5 0
3 years ago
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