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NeTakaya
3 years ago
7

What major problems could you encounter in complex intersections?

Engineering
1 answer:
Natali5045456 [20]3 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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What historical event allowed both aerospace fields to make enormous strides<br> forward? *
galina1969 [7]

Answer:

The world wars. Most notably world War II

Explanation:

The demand for aircrafts during these events led to extensive research into the design of aircrafts. Aircraft advanced within these years from a simple design to a more complex design; capable of carrying fire power and even became bomb equipped. Also, the material of choice of production moved from wood to metal and the engine was improved on to gain more speed and maneuverability.

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When asked about favorite Thanksgiving leftovers, 9% of the people said turkey and 7100 said mashed potatoes. Which food is more
irakobra [83]

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how many people were asked though

Explanation:

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3 years ago
A good way of ensuring whether a listener has understood a presentation is to record all major and minor details. identify all u
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Summarize in one's own words

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4 years ago
Read 2 more answers
Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series
love history [14]

Answer:

The required pumping head is 1344.55 m and the pumping power is 236.96 kW

Explanation:

The energy equation is equal to:

\frac{P_{1} }{\gamma } +\frac{V_{1}^{2}  }{2g} +z_{1} =\frac{P_{2} }{\gamma } +\frac{V_{2}^{2}  }{2g} +z_{2}+h_{i} -h_{pump} , if V_{1} =0,z_{2} =0\\h_{pump} =\frac{V_{2}^{2}}{2} +h_{i}-z_{1}

For the pipe 1, the flow velocity is:

V_{1} =\frac{Q}{\frac{\pi D^{2} }{4} }

Q = 18 L/s = 0.018 m³/s

D = 6 cm = 0.06 m

V_{1} =\frac{0.018}{\frac{\pi *0.06^{2} }{4} } =6.366m/s

The Reynold´s number is:

Re=\frac{\rho *V*D}{u} =\frac{999.1*6.366*0.06}{1.138x10^{-3} } =335339.4

\frac{\epsilon }{D} =\frac{0.00026}{0.06} =0.0043

Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941

The head of pipe 1 is:

h_{1} =\frac{V_{1}^{2}  }{2g} (k_{L}+\frac{fL}{D}  )=\frac{6.366^{2} }{2*9.8} *(0.5+\frac{0.0294*20}{0.06} )=21.3m

For the pipe 2, the flow velocity is:

V_{2} =\frac{0.018}{\frac{\pi *0.03^{2} }{4} } =25.46m/s

The Reynold´s number is:

Re=\frac{\rho *V*D}{u} =\frac{999.1*25.46*0.03}{1.138x10^{-3} } =670573.4

\frac{\epsilon }{D} =\frac{0.00026}{0.03} =0.0087

The head of pipe 1 is:

h_{2} =\frac{V_{2}^{2}  }{2g} (k_{L}+\frac{fL}{D}  )=\frac{25.46^{2} }{2*9.8} *(0.5+\frac{0.033*36}{0.03} )=1326.18m

The total head is:

hi = 1326.18 + 21.3 = 1347.48 m

The required pump head is:

h_{pump} =\frac{25.46^{2} }{2*9.8} +1347.48-36=1344.55m

The required pumping power is:

P=Q\rho *g*h_{pump}  =0.018*999.1*9.8*1344.55=236965.16W=236.96kW

8 0
3 years ago
How do you distinguish the difference between drag and lift forces?
kumpel [21]

Answer Explanation :

LIFT FORCE:

  • Lift force is that force which acts right angles to the direction of motion
  • Lift forces is produced through air
  • Lift forces is created by difference in air pressure
  • force on wings of bird is an example of lift force

DRAG FORCE:

  • Drag force is that force which act opposite to the direction of motion
  • Drag is caused by friction and differences in air pressure
  • Skydiving is an example of drag force

6 0
4 years ago
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