What is the primary damage control telephone circuit for

Ben İngiliz oldum düzelte bilirmiyim

Lelu [443]

Answer:

What laguange is that?

Explanation:

7 0

3 years ago

After testing a model of a fuel-efficient vehicle, scientists build a full-sized vehicle with improved fuel efficiency. Which st

Mazyrski [523]

B evaluate the solution. Sorry if I'm wrong.

4 0

3 years ago

What does a manufacturing engineer do?

Lisa [10]

Answer: Manufacturing engineers may be responsible for solving production problems, conducting cost-benefit assessments, or designing and manufacturing goods and systems using computer-aided design software. Plant engineers and process engineers are two terms used by professionals in this industry. Manufacturing engineers are in charge of new and existing production lines' technical management, maintenance, and development. Employers are looking for people that are commercially aware and have good technical and analytical skills. They are in charge of keeping production costs low while preserving the product or service's quality, and they have considerable project expertise and insight.

Explanation: See above.

I hope this helps.

6 0

2 years ago

Read 2 more answers

6.28 A six-lane freeway (three lanes in each direction) in rolling terrain has 10-ft lanes and obstructions 4 ft from the right

dimulka [17.4K]

Answer:

Assume Base free flow speed (BFFS) = 70 mph

Lane width = 10 ft

Reduction in speed corresponding to lane width, fLW = 6.6 mph

Lateral Clearance = 4 ft

Reduction in speed corresponding to lateral clearance, fLC = 0.8 mph

Interchanges/Ramps = 9/ 6 miles = 1.5 /mile

Reduction in speed corresponding to Interchanges/ramps, fID = 5 mph

No. of lanes = 3

Reduction in speed corresponding to number of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 70 – 6.6 – 0.8 – 3 – 5 = 54.6 mph

Peak Flow, V = 2000 veh/hr

Peak 15-min flow = 600 veh

Peak-hour factor = 2000/ (4*600) = 0.83

Trucks and Buses = 12 %

RVs = 6 %

Rolling Terrain

fHV = 1/ (1 + 0.12 (2.5-1) + 0.06 (2.0-1)) = 1/1.24 = 0.806

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 2000/ (0.83*3*0.806*1.0) = 996.54 ~ 997 veh/hr/ln

Vp < (3400 – 30 FFS)

S = FFS

S = 54.6 mph

Density = Vp/S = (997) / (54.6) = 18.26 veh/mi/ln

7 0

3 years ago

. The flexure strength test was performed on a concrete beam having a cross section of 0.15m by 0.15m and a span of 0.45m. If th

ivann1987 [24]

Answer:

σ =5.39Mpa

Explanation:

step one:

The flexure strength is defined as the tendency with which unreinforced concrete yield to bending forces

Flexural strength test Flexural strength is calculated using the equation:

σ = FL/ (bd^2 )----------1

Where

σ = Flexural strength of concrete in Mpa

F= Failure load (in N).

L= Effective span of the beam

b= Breadth of the beam

step two:

Given data

F=40.45 kN= 40450N

b=0.15m

d=0.15m

L=0.45m

step three:

substituting into the expression we have

σ = 40450*0.45/ (0.15*0.15^2 )

σ =18202.5/ (0.15*0.15^2 )

σ =18202.5/ (0.15*0.0225 )

σ =18202.5/0.003375

σ =5393333.3

σ =5393333.3/1000000

σ =5.39Mpa

Therefore the flexure strength of the concrete is 5.39Mpa

5 0

3 years ago