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Alja [10]
3 years ago
12

A segment of a roadway has a free flow speed of 45 mph and a jam density of 25 ft per vehicle. Determine the maximum flow and at

a flow rate of 1950 vph
Engineering
1 answer:
-BARSIC- [3]3 years ago
4 0

Answer:

2376 vph

Explanation:

Given data

free flow speed ( Vf ) = 45 mph

Jam density = 25 ft per vehicle

flow rate = 1950 vph

first we calculate the Jam density in vehicle /mile

= 5280 / 25 = 211.2 vehicle/mile

where ; 1 mile = 5280 feet

The maximum flow can be calculated using Greenshield method

q = ( Vf * jam density ) / 4  =  ( 45 * 211.2 ) / 4

  = 2376 vph

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Answer:

reasearching the problem

Explanation:

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a

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common sense

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3 years ago
Lithium at 20°C is BCC and has a lattice constant of 0.35092 nm. Calculate a value for the atomic radius of a lithium atom in na
Nutka1998 [239]

Answer:

the atomic radius of a lithium atom is 0.152 nm

Explanation:

Given data in question

structure = BCC

lattice constant  (a) = 0.35092 nm

to find out

atomic radius of a lithium atom

solution

we know structure is BCC

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here we have known a value so we put a in radius formula

radius =  \sqrt{3} /4 × a

radius =  \sqrt{3} /4 × 0.35092

radius = 0.152 nm

so the atomic radius of a lithium atom is 0.152 nm

5 0
3 years ago
What is the weight density of a 2.24 in diameter titanium sphere that weights 0.82 lb?
nasty-shy [4]

Answer:

0.14\ lb/in^{3}

Explanation:

Density is defined as mass ler unit volume, expressed as

\rho=\frac {m}{v}

Where m is mass, \rho is density and v is the volume. For a sphere, volume is given as

v=\frac {4\pi r^{3}}{3}

Replacing this into the formula of density then

\rho=\frac {m}{\frac {4\pi r^{3}}{3}}

Given diameter of 2.24 in then the radius is 1.12 in. Substituting 0.82 lb for m then

\rho=\frac {0.82}{\frac {4\pi 1.12^{3}}{3}}=0.13932044952427\approx 0.14 lb/in^{3}

4 0
4 years ago
when removing the pistons and rods assemblies from a cylinder block technician a positions the throw of the crankshaft at the to
defon

Answer: Technician A

Reason: Piston and rod are connected with crank shaft and connecting rod. Smaller end of connecting rod is connected with piston and bigger end is connected with crankshaft.  

<u>Explanation: </u>

When removing the piston and rod assemblies from a cylinder block: The technician with correct approach. So to remove piston from cylinder technician must throw crankshaft first and the remove connecting rod by losing nuts and caps...So technician A is in right way. Technician B using hammer to remove piston from the rod not possible because connecting rod and piston connected by nuts and caps can’t be separate by hammer.  

Technician A positions the throw of the crankshaft at the top of its stroke and removes the connecting rod nuts and cap.

Technician B covers the rod bolts with hammers and pushes the piston and rod assembly out with the wooden hammer handle or wooden drift and supports the piston as it comes out of the cylinder.

4 0
3 years ago
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