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2 years ago

A _________ can used for both entering and exiting a highway.

Engineering

1 answer:

Helga [31]2 years ago

7 0

The lane that can be used for both entering and exiting a highway is called; A weave lane

<h3>How to enter a highway?</h3>

When it comes to road highways, there are different terminologies that apply such as Acceleration lane, Deceleration Lane, Interchange, Weave lane, Urban roadway e.t.c

However, when it comes to the one that can be used for both entering and exiting a highway, the correct answer will be Weave Lane because it has both an entrance and an exit for an expressway. Traffic may come onto and leave the expressway at the same location.

Read more about road highway at; brainly.com/question/12961781

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A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 78 MPa (70.98 ksi). If the plate is

Reika [66]

Answer:

minimum length of a surface crack is 15.043 mm

Explanation:

given data

strain fracture toughness K = 78 MPa

tensile stress = 345 MPa

Y = 1.04

to find out

minimum length of a surface crack

solution

we find here length of critical interior flaw from formula that is

α = $\frac{1}{\pi} (\frac{K}{\sigma Y})^2$ ....................1

put here value we get

α = $\frac{1}{\pi} (\frac{78*\sqrt{10^3} }{345*1.04})^2$

α = 15.043 mm

so minimum length of a surface crack is 15.043 mm

7 0

3 years ago

A tank with a volume of 8 m3 containing 4 m3 of 20% (by volume) NaOH solution is to be purged by adding pure water at a rate of

lawyer [7]

Answer:

The time necessary to purge 95% of the NaOH is 0.38 h

Explanation:

Given:

vfpure water(i) = 3 m³/h

vNaOH = 4 m³

xNaOH = 0.2

vfpure water(f) = 2 m³/h

pwater = 1000 kg/m³

pNaOH = 1220 kg/m³

The mass flow rate of the water is = 3 * 1000 = 3000 kg/h

The mass of NaOH in the solution is = 0.2 * 4 * 1220 = 976 kg

When the 95% of the NaOH is purged, thus the NaOH in outlet is = 0.95 * 976 = 927.2 kg

The volume of NaOH in outlet after time is = 927.2/1220 = 0.76 m³

The time required to purge the 95% of the NaOH is = 0.76/2 = 0.38 h

4 0

4 years ago

technician a says that the higher the gear selected the more torque is available. technician b says that a transaxle contains ge

umka21 [38]

Technician A is correct. Technician B is wrong because a gear's transmission is used to increase or decrease torque.

The relation torque is relying on multiplying the circumferential detail with the resource of the usage of the radius; massive gears experience a greater amount of torque, at the same time as smaller gears experience a great deal much less torque. Similarly, the torque ratio is equal to the ratio of the gears' radii. A gear's transmission torque modifications as it will boom or decreases speed. Commonly, with the resource of the usage of lowering the speed, a small torque on the doorway issue is transferred as a massive torque at the output issue. The calculation of torque is quantified with the resource of the usage of an extensive form of teeth.

Learn more about the torque at brainly.com/question/28220969

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3 0

1 year ago

A small metal particle passes downward through a fluid medium while being subjected to the attraction of a magnetic field such t

bekas [8.4K]

Answer:

a)Δs = 834 mm

b)V=1122 mm/s

$a=450\ mm/s^2$

Explanation:

Given that

$s = 15t^3 - 3t\ mm$

When t= 2 s

$s = 15t^3 - 3t\ mm$

$s = 15\times 2^3 - 3\times 2\ mm$

s= 114 mm

At t= 4 s

$s = 15t^3 - 3t\ mm$

$s = 15\times 4^3- 3\times 4\ mm$

s= 948 mm

So the displacement between 2 s to 4 s

Δs = 948 - 114 mm

Δs = 834 mm

We know that velocity V

$V=\dfrac{ds}{dt}$

$\dfrac{ds}{dt}=45t^2-3$

At t= 5 s

$V=45t^2-3$

$V=45\times 5^2-3$

V=1122 mm/s

We know that acceleration a

$a=\dfrac{d^2s}{dt^2}$

$\dfrac{d^2s}{dt^2}=90t$

a= 90 t

a = 90 x 5

$a=450\ mm/s^2$

4 0

3 years ago

A 600 MW power plant has an efficiency of 36 percent with 15

ololo11 [35]

Answer:

401.3 kg/s

Explanation:

The power plant has an efficiency of 36%. This means 64% of the heat form the source (q1) will become waste heat. Of the waste heat, 85% will be taken away by water (qw).

qw = 0.85 * q2

q2 = 0.64 * q1

p = 0.36 * q1

q1 = p /0.36

q2 = 0.64/0.36 * p

qw = 0.85 *0.64/0.36 * p

qw = 0.85 *0.64/0.36 * 600 = 907 MW

In evaporation water becomes vapor absorbing heat without going to the boiling point (similar to how sweating takes heat from the human body)

The latent heat for the vaporization of water is:

SLH = 2.26 MJ/kg

So, to dissipate 907 MW

G = qw * SLH = 907 / 2.26 = 401.3 kg/s

8 0

4 years ago

Read 2 more answers