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just olya [345]
4 years ago
13

What is the fundamental relationship among speed (u), density (k), and flow (q)?

Physics
1 answer:
Semenov [28]4 years ago
3 0

The representation of the traffic flow is determined by three main characteristics of this: The flow, the density and the speed. The relationship between the three helps to improve the design of the roads as well as all kinds of facilities operations.

The mathematical relationship that groups them is given by the function:

q = k*\nu

Where,

\nu= Speed

k = Density

q = Flow

Through it it is possible to obtain properties of this by which,

  • When the density on the road is zero, the flow is also zero because there are no vehicles on the road
  • As the density increases, the flow increases
  • When the density reaches a maximum jam density, the flow must be zero because the vehicles will line up from end to end
  • The flow will also increase to a maximum value, increases in density beyond that point result in flow reductions.
  • Speed is the average speed of space.

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Answer:

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Explanation:

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Explanation:

Given that,

Length of the spring, l = 50 cm = 0.5 m

Mass connected to the end, m = 330 g = 0.33 kg

The mass is released and falls, stretching the spring by 30 cm before coming to rest at its lowest point. On applying Newton's second law, 10 cm below the release point, x = 15 cm

(a) When the mass is connected, the force of gravity is balanced by the force in spring.

kx=mg\\\\k=\dfrac{mg}{x}\\\\k=\dfrac{0.33\times 9.8}{0.15}\\\\k=21.56\ N/m

(b) The amplitude of the oscillation will be 15 cm as it is half of the total distance travelled.

(c) The frequency of the oscillation is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{21.56}{0.33}} \\\\f=1.28\ Hz

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When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the
Taya2010 [7]

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

\theta = 54.6^{\circ}

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = \sqrt{\frac{gx}{sin2\theta}}

v = \sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}

v = \sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}

v = \sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s

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The angular velocity can be calculated as:

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