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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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A rich industrialist was found murdered in his house. The police arrived at the scene at 11:00 PM. The temperature of the corpse
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The dude was killed around 6:30PM

Explanation:

Newton's law of cooling states:

    T = T_m + (T_0-T_m)e^{kt}

where,

T_0 = initial temp

T_m = temp of room

T = temp after t hours

k = how fast the temp is changing

t = time (hours)

T_0 = 31     because the body was initlally 31ºC when the police found it

T_m = 22   because that was the room temp

T = 30  because the body temp drop to 30ºC after 1 hour

t = 1 because that's the time it took for the body temp to drop to 30ºC

k=???   we don't know k so we must solve for this

rearrange the equation to solve for k

T = T_m + (T_0-T_m)e^{kt}

T - T_m= (T_0-T_m)e^{kt}

\frac{T - T_m}{(T_0-T_m)}= e^{kt}

ln(\frac{T - T_m}{T_0-T_m})=kt

\frac{ln(\frac{T - T_m}{T_0-T_m})}{t}=k

plug in the numbers to solve for k

k = \frac{ln(\frac{T - T_m}{T_0-T_m})}{t}

k = \frac{ln(\frac{30 - 22}{31-22})}{1}

k=ln(\frac{8}{9})

Now that we know the value for k, we can find the moment the murder occur. A crucial information that the question left out is the temperature of a human body when they're still alive. A living human body is about 37ºC. We can use that as out initial temperature to solve this problem because we can assume that the freshly killed body will be around 37ºC.

T_0 = 37     because the body was 37ºC right after being killed

T_m = 22   because that was the room temp

T = 31  because the body temp when the police found it

k=ln(\frac{8}{9})   we solved this earlier

t = ???   we don't know how long it took from the time of the murder to when the police found the body

Rearrange the equation to solve for t

T = T_m + (T_0-T_m)e^{kt}

T - T_m= (T_0-T_m)e^{kt}

\frac{T - T_m}{(T_0-T_m)}= e^{kt}

ln(\frac{T - T_m}{T_0-T_m})=kt

\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}=t

plug in the values

t=\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}

t=\frac{ln(\frac{31 - 22}{37-22})}{ln(8/9)}

t=\frac{ln(3/5)}{ln(8/9)}

t=\frac{ln(3/5)}{ln(8/9)}

t ≈ 4.337 hours from the time the body was killed to when the police found it.

The police found the body at 11:00PM so subtract 4.337 from that.

11 - 4.33 = 6.66 ≈ 6:30PM

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