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2 years ago

What major problems could you encounter in complex intersections?

Engineering

1 answer:

Natali5045456 [20]2 years ago

3 0

Types of Problems

Inappropriate intersection traffic control.

Inadequate visibility of the intersection or regulatory traffic control devices.

Inadequate intersection sight distance.

Inadequate guidance for motorists.

Excessive intersection conflicts within or near the intersection.

Vehicle conflicts with non-motorists.

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Scientific research techniques are used to analyze the effectiveness of political advertising. False True

miv72 [106K]

Answer:

correct me if i'm wrong but i think it's false

Explanation:

5 0

3 years ago

Bore = 3"

Grace [21]

I need help my self lol XD

5 0

3 years ago

Read 2 more answers

A furnace wall is to be built of 20-cm firebrick and building (structural) brick of same thickness. The thermal conductivities o

Norma-Jean [14]

Answer:

q=2313.04 $W/m^2$

T=690.86°C

Explanation:

Given that

Thickness t= 20 cm

Thermal conductivity of firebrick= 1.6 W/m.K

Thermal conductivity of structural brick= 0.7 W/m.K

Inner temperature of firebrick=980°C

Outer temperature of structural brick =30°C

We know that thermal resistance

$R=\dfrac{t}{KA}$

These are connect in series

$R=\left(\dfrac{t}{KA}\right)_{fire}+\left(\dfrac{t}{KA}\right)_{struc}$

$R=\dfrac{0.2}{1.6A}+\dfrac{0.2}{0.7A}\ K/W$

$R=\dfrac{23}{56A}\ K/W$

Heat transfer

$Q=\dfrac{\Delta T}{R}$

$Q=56A\times \dfrac{980-30}{23}\ W$

So heat flux

q=2313.04 $W/m^2$

Lets temperature between interface is T

Now by equating heat in both bricks

$\dfrac{980-T}{\dfrac{0.2}{1.6A}}=\dfrac{T-30}{\dfrac{0.2}{0.7A}}$

So T=690.86°C

6 0

3 years ago

How many GT2RS cars were made in 2019

labwork [276]

Answer:

1000

Explanation:

3 0

3 years ago

Read 2 more answers

t is desired to produce aligned carbon fiber-epoxy matric composite having a longitudinal tensile strength of 800 MPa. Given (a)

Nesterboy [21]

Answer:

The value of critical length = 3.46 mm

The value of volume of fraction of fibers = 0.43

Explanation:

Given data

$\sigma_{T}$ = 800 M pa

D = 0.017 mm

L = 2.3 mm

$\sigma_{f}$ = 5500 M pa

$\sigma_{m}$ = 18 M pa

$\sigma_c$ = 13.5 M pa

(a) Critical fiber length is given by

$L_{c} = \sigma_{f} (\frac{D}{2 \sigma_{c} } )$

Put all the values in above equation we get

$L_{c} =5500 (\frac{0.017}{(2) (13.5)} )$

$L_{c} = 3.46$ mm

This is the value of critical length.

(b).Since this critical length is greater than fiber length Than the volume fraction of fibers is given by

$V_{f} = \frac{\sigma_T - \sigma_m}{\frac{L\sigma_c}{D} - \sigma_m }$

Put all the values in above formula we get

$V_{f} = \frac{800-18}{\frac{(2.3)(13.5)}{0.017} - 18 }$

$V_{f}$ = 0.43

This is the value of volume of fraction of fibers.

3 0

3 years ago