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3 years ago

Find the value of L

Engineering

1 answer:

KonstantinChe [14]3 years ago

6 0

Answer:

the value is 356732 Volt

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Why is concrete on its own not a good material to use

Inga [223]

Answer:

It has poor tensile strength despite having high compressive strength

Explanation:

Concrete exhibits high compressive strength when used. However, it has very low compressive strength. This is the reason why concrete is normally combined with steel to make a composite building material called reinforced concrete. The steel reinforces concrete hence increasing the tensile strength in RC buildings. The end composite is durable and fireproof. Generally, the main reason why concrete is not use on its own is due to its poor tensile strength.

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3 years ago

Say that a variable A in CFG G is necessary if it appears in every derivation of some string w ∈ G. Let NECESSARY CFG = {hG, Ai|

ale4655 [162]

Answer:

Explanation:

solution

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3 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

I m from India<br><br><br>5603642259 pd 123456

Volgvan

Answer:

Hi there...

Explanation:

Please give me brainliest :)

6 0

3 years ago

Read 2 more answers

For the given network, if R = 0.8 [Ω], the natural response of i in(t) is:

11111nata11111 [884]

Answer:

Iin(t) =1.3 × (-0.8/1.56) e^-t/1.56 A

Explanation:

In physics, the determination of the term " natural response" simply means that we want to know what happens in a circuit when the value of t = 0, that is to say after the circuit has been disconnected. Hence, the value of the voltage and the current can then be determined or Calculated;

For the the natural response of i in(t) we will be using the formula below;

I(t) = Vo × t/ R = Vo/R × e^-t/h.

Where h = 1/RC = time constant.

For t= 0^- = 0.8 × 1= 0.8 V.

1/Ctotal = 1/ 2 + 1/3 = 6/5

For t = 0^+;

h =( 0.8 + 0.5) × 6/5 = 1.56 seconds.

Hence, we will have;

Vin(t) = 0.8 × e^-t/1.56.

Iin(t) =1.3 × (-0.8/1.56) e^-t/1.56

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3 years ago