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2 years ago

7

What major problems could you encounter in complex intersections?

1 answer:

Natali5045456 [20]2 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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Answer:

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Explanation:

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3 years ago

A rectangular channel 2 m wide carries 3 m3 /s of water at a depth of 1.2 m. If an obstruction 40 cm wide is placed in the middl

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8 0

3 years ago

Find E[x] when x is sum of two fair dice?

Ksenya-84 [330]

Answer:

When two fair dice are rolled, 6×6=36 observations are obtained.

P(X=2)=P(1,1)=

36

1

P(X=3)=P(1,2)+P(2,1)=

36

2

=

18

1

P(X=4)=P(1,3)+P(2,2)+P(3,1)=

36

3

=

12

1

P(X=5)=P(1,4)+P(2,3)+P(3,2)+P(4,1)=

36

4

=

9

1

P(X=6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)=

36

5

P(X=7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)+P(6,1)=

36

6

=

6

1

P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+P(6,2)=

36

5

P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3)=

36

4

=

9

1

P(X=10)=P(4,6)+P(5,5)+P(6,4)=

36

3

=

12

1

P(X=11)=P(5,6)+P(6,5)=

36

2

=

18

1

P(X=12)=P(6,6)=

36

1

Therefore, the required probability distribution is as follows.

Then, E(X)=∑X

i

⋅P(X

i

)

=2×

36

1

+3×

18

1

+4×

12

1

+5×

9

1

+6×

36

5

+7×

6

1

+8×

36

5

+9×

9

1

+10×

12

1

+11×

18

1

+12×

36

1

=

18

1

+

6

1

+

3

1

+

9

5

+

6

5

+

6

7

+

9

10

+1+

6

5

+

18

11

+

3

1

=7

E(X

2

)=∑X

i

2

⋅P(X

i

)

=4×

36

1

+9×

18

1

+16×

12

1

+25×

9

1

+36×

36

5

+49×

6

1

+64×

36

5

+81×

9

1

+100×

12

1

+121×

18

1

+144×

36

1

=

9

1

+

2

1

+

3

4

+

9

25

+5+

6

49

+

9

80

+9+

3

25

+

18

121

+4

=

18

987

=

6

329

=54.833

Then, Var(X)=E(X

2

)−[E(X)]

2

=54.833−(7)

2

=54.833−49

=5.833

∴ Standard deviation =

Var(X)

=

5.833

=2.415

4 0

2 years ago

An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p

Veseljchak [2.6K]

Answer:

Ts =Ta E)- 300(

569.5 K

5

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

10

24.9x10 m

=

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

14

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

5

(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

AH

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

J

mol

AUs¡ = Ws¡ = -5601.55

3 0

3 years ago