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3 years ago

Question Completion Status:

Engineering

1 answer:

Mashutka [201]3 years ago

6 0

Answer: c. Centre of pressure

Explanation:

Pressure is applied on a surface when a force is exerted on a particular point on that surface by another object when the two come into contact with each other.

The point where the pressure is applied is known as the centre of the pressure with the force then spreading out from this point much like an epicentre in an earthquake.

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A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the

Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + $\frac{u^2}{30(\frac{a}{g}\pm G)}$ ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5 + $\frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}$

solve it we get

SSD = 644 ft

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1 - $cos (\frac{28.65 SSD}{Rv})$ ) .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1 - $cos (\frac{28.65 \times 664}{2294})$ )

solve it we get

Ms = 22.57 ft

and

superelevation rate for the curve will be here as

R = $\frac{u^2}{15(e+f)}$ ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = $\frac{65^2}{15(e+0.10)}$

solve it we get

e = 2%

3 0

3 years ago

Technician A says the final drive assembly always has a gear ratio of 1:1. Technician B says the final drive assembly provides f

Olenka [21]

Answer:

Technician B only is correct

Explanation:

The last stage of gears found between the vehicle transmission system and the wheels is the final drive ratio. The function of the final drive gear assembly is to enable a gear reduction control stage to reduce the rotation per minute and increase the wheel torque, such that the vehicle performance can be adjusted and the final gear ratio can be between 3:1 and 4.5:1 not 1:1

Therefore, technician B only is correct

5 0

4 years ago

1. There are two categories (shapes) for the Virginia driver's license. The ________________________ shape license represents th

dolphi86 [110]

Answer:

Vertical; horizontal.

Explanation:

The Virginia Department of Motor Vehicles started issuing sets of newly designed driver's licenses to drivers in 2009. Although, the cards that were issued to drivers prior to the introduction of the new cards remained valid until they were expired.

There are two categories (shapes) for the Virginia driver's license. The vertical shape license represents the driver who is under the age of 21, and the horizontal shaped license represents the driver who is over the age of 21.

Additionally, the Virginia's driver license (vertical in shape) issued to drivers who are under the age of 21 has a background image of a dogwood flower while the horizontal shaped license issued to drivers who are over the age of 21 has a background image of the state capitol.

5 0

3 years ago

A spherical container made of steel has 20 ft outer diameter and wal thickness of 1/2 inch. Knowing the internal pressure is 50

anastassius [24]

Answer:

maximum normal stress = 5975 psi

maximum shear stress = 2987.50 psi

Explanation:

Given data

dia = 20 ft

wall thickness = 1/2 inch

internal pressure = 50 psi

To find out

the maximum normal stress and the maximum shearing stress

Solution

By the Mohr's circle we will find out shear stress

first we calculate inner radius

i.e. r = (diameter/2) - t

r = (20 × 12 in )/2 - ( 1/2 )

r = 120 - 0.5 = 119.5 inch

Now we find out maximum normal stress by given formula

normal stress = ( internal pressure× r ) / 2 t

normal stress = ( 50×119.5 ) / 2 × 0.5

maximum normal stress = 5975 psi

and minimum normal stress is 0, due to very small radius

and maximum shear stress will be

shear stress = ( maximum normal stress - minimum normal stress ) / 2

shear stress = ( 5975- 0 ) / 2

maximum shear stress = 2987.50 psi

5 0

4 years ago

How many trips would one rubber-tired Herrywampus have to make to backfill a space with a geometrical volume of 5400 cubic yard?

nikklg [1K]

Answer:

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

Explanation:

a) By volume.

The shrinkage factor is:

$\frac{5400cu-yd}{1-0.25} =7200cu-yd$

The volume at loose is:

$V_{loose} =V_{bank} (1+swell-factor)=7200(1+0.2)=8640cu-yd$

If the Herrywampus has a capacity of 30 cubic yard:

$\frac{8640cu-yd}{30cu-yd/trip} =288trip$

b) By weight

The swell factor in terms of percent swell is equal to:

$pounds-per-cubic-yard-loose=\frac{pounds-per-cubic-yard-bank}{\frac{percent-swell}{100}+1 }$

$pounds-per-cubic-yard-loose=\frac{3000}{\frac{20}{100} +1} =2500lb/cu-yd$

The weight of backfill is:

$8640cu-yd*2500\frac{lb}{cu-yd} *\frac{1ton}{2000lb} =10800ton$

The Herrywampus has a capacity of 40 ton:

$\frac{10800}{40ton/trip} =270trip$

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

8 0

3 years ago