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Anna35 [415]
3 years ago
12

You are observing traffic in a single lane of a highway at a specific location. You measure the average headway and average spac

ing of passing vehicles as 3 seconds and 150 feet, respectively. Calculate the flow, speed and density of the traffic stream in this lane.
Physics
1 answer:
HACTEHA [7]3 years ago
6 0

Answers:

1) flow of traffic =   1198.8 veh/h

2) average speed = 34.09 mi/h

3)density of traffic = 34.34 veh/mi

Explanation:

1) to find flow of traffic

we use the relation

q=1/h

where q is the traffic flow and h is average time headway.

h= 3s       (given)

insert the value

q=1/3=0.333veh/s=1198.8veh/h

2) to find average traffic speed

use the relation

u=S/h

where u is average speed S is average spacing and h average time

S= 150 ft   (given)

so inserting the values

u= 150/3*3600/5280=34.09 mi/h

3) density of traffic

K=q/u

where K is density of traffic q is flow of traffic and u is average speed

inserting values from above solved parts

K=1198.8/34.09=34.34 veh/mi

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Anvisha [2.4K]

Answer:

Weight is what you get when a certain amount of gravity is acting on that mass, and something, like the surface of a planet, is resisting that action. In space, when falling freely, there's nothing resisting the pull of gravity so weight disappears. Mass however stays.

hope this helps u

Explanation:

7 0
3 years ago
A rock falls off the edge of a building and falls for 10 seconds. What was
solmaris [256]
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If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front o
andrezito [222]
The answer is

18 / x = 12 / 4 
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3 years ago
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______ involves organizing and breaking down information into easier groups to expand capacity. Rehearsal is the verbal repetiti
adell [148]

Hi!


The answers would be <u>chunking</u> & <u>short-term</u>


1. <u>Chunking </u>involves organizing and breaking down information into easier groups to expand capacity.

<h3>Explanation:</h3>

Chunking is a mental process that is observed to increase short-term memory by taking the information and categorizing it into small groups. For instance, a longer number taken as a single unit is harder to recall then when it is divided into smaller units. 235469350 is harder to instantly recall as compared to when it is chunked into 3 groups: 235 469 350.

This allows more information to be stored in, thereby increasing the capacity of the mind to store information.


2. Rehearsal is the verbal repetition of information. These techniques are especially important for the improvement of <u>short-term</u> memory.

<h3>Explanation: </h3>

Short-term memory is lost after a couple of seconds or minutes, for instance even if you chunk the information, you might not recall it after 30 seconds. Rehearsing or repetition of information, either loudly or mentally, extends the time a particular information is retained.

So you depending on the number of times you repeat the number 235 469 350, the more your short term memory improves .


Hope this helps!


6 0
3 years ago
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0
3 years ago
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