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Hatshy [7]
2 years ago
5

A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the

capacity of the highway section, the speed at capacity, and the density when the highway is at one-quarter of its capacity?
Physics
1 answer:
lions [1.4K]2 years ago
3 0

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

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Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

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A quantity y is to be determined from the equation y=(px)/q^2
ki77a [65]

Answer:

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A person lifting a chair is converting chemical energy to mechanical energy.
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Answer:

630cm/s

Explanation:

In simple harmonic motion, the tangential velocity is expressed mathematically as v = ὦr

ὦ is the angular velocity = 2πf

r is the radius of the disk

f is the frequency

Given the radius of disk = 10cm

frequency = 10Hz

v = 2πfr

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v = 628.32 cm/s

The tangential velocity = 630cm/s ( to 2 significant figures)

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What visible signs indicate a precipitation reaction when two solutions are mixed?
Illusion [34]

Formation of an insoluble solid

Explanation:

One of the remarkable visible signs that indicates a precipitation reaction when two solutions are mixed is the formation of an insoluble solid. The insoluble solid formed is the precipitate.

  • Precipitates usually forms in single replacement reactions and double replacement or double decomposition reactions.
  • They form when two soluble compounds react. One of the product is an insoluble solid in the solution called the precipitate.
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6 0
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