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lorasvet [3.4K]
1 year ago
9

The maximum capacity of a 2-lane carriageway of a four lane dual carriageway is 2000 veh/hour. due to pipe laying operations the

width of two lane carriageway is reduced, restricting the maximum capacity to 1100 veh/hour. when the flow upstream beyond the influence of the bottleneck is reasonably steady and freeflowing at 1500 veh/hour. find i) the mean speed of traffic in the bottleneck ii) the rate at which the queue of the congested conditions outside bottleneck grows. the mean space headway when the vehicles are stationary is 8 m. the relation between speed and concentration is linear
Engineering
1 answer:
Yuki888 [10]1 year ago
3 0

Based on the maximum capacity as a result of the pipe-laying operations, and the maximum capacity without obstruction to the four-lane dual carriageway, the mean speed of traffic in the bottleneck is -2.73km.

<h3>What is the mean speed of traffic?</h3>

This can be found as:

= (Maximum restricted capacity - freeflowing rate) / (Kb - Ka)

= (1,100 - 1,500) / (209 - 62.5)

= -2.73 km/h

The rate that the queue outside the bottleneck grows is:

= Free flowing rate - (mean speed of traffic x ka)

= 1,500 - (-2.73 x 62.5)

= 1,670 veh/hour

Find out more on bottleneck at brainly.com/question/9551615

#SPJ1

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nirvana33 [79]

Answer:

Q(h=200)=0.35W

Q(h=3000)=5.25W

Explanation:

first part h=200W/Km^2

we must use the convection heat transfer equation for the chip

Q=hA(Ts-T∞)

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A=Base*Leght=5mmx5mm=25mm^2

Ts=temperature of the chip=85C

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Q=200x2.5x10^-5(85-15)=0.35W

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Q=3000x2.5x10^-5(85-15)=5.25W

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3 years ago
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If x &lt; 5 and x &gt;c, give a value of c such that there
Arlecino [84]

we have  

x<5

x>c

we know that

The solution is the intersection of both solution sets of the given inequalities.  

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Because

A number cannot be both less than 5 and greater than 5 at the same time

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A battery with an f.e.m. of 12 V and negligible internal resistance is connected to a resistor of 545 How much energy is dissipa
Alenkasestr [34]

Answer:

When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then  

R

1

 in Figure 1(a) could be the resistance of the screwdriver’s shaft,  

R

2

 the resistance of its handle,  

R

3

 the person’s body resistance, and  

R

4

 the resistance of her shoes.

Figure 2 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.)

Two electrical circuits are compared. The first one has three resistors, R sub one, R sub two, and R sub three, connected in series with a voltage source V to form a closed circuit. The first circuit is equivalent to the second circuit, which has a single resistor R sub s connected to a voltage source V. Both circuits carry a current I, which starts from the positive end of the voltage source and moves in a clockwise direction around the circuit.

Figure 2. Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right).

To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 2.

According to Ohm’s law, the voltage drop,  

V

, across a resistor when a current flows through it is calculated using the equation  

V

=

I

R

, where  

I

 equals the current in amps (A) and  

R

 is the resistance in ohms  

(

Ω

)

. Another way to think of this is that  

V

 is the voltage necessary to make a current  

I

 flow through a resistance  

R

.

So the voltage drop across  

R

1

 is  

V

1

=

I

R

1

, that across  

R

2

 is  

V

2

=

I

R

2

, and that across  

R

3

 is  

V

3

=

I

R

3

. The sum of these voltages equals the voltage output of the source; that is,

V

=

V

1

+

V

2

+

V

3

.

 

This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation  

P

E

=

q

V

, where  

q

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V

 is the voltage. Thus the energy supplied by the source is  

q

V

, while that dissipated by the resistors is

q

V

1

+

q

V

2

+

q

V

3

.

Explanation:

8 0
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