Some cities now implement signal lights designed to specifically apply to _____ rather than motorized vehicles.

Physics

1 answer:

dimulka [17.4K]3 years ago

7 0

The correct answer is C; Bicycles.

Further Explanation:

In major cities, in the United States, have implemented signal lights specifically designed for bicycle riders. The riders also have their own designated bike lanes in many large cities. Drivers in vehicles, are to give the right of way to people on bicycles.

Bicycle riders are to follow the same laws and laws specifically for the riders or they can face fines and tickets.

Learn more about bicycle laws at brainly.com/question/8934107

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A 90kg mountain climber hangs from a nylon rope and streches it ny 25.0cm. if the rope was originally 30.0m long and it's diamet

bixtya [17]

Answer:

Explanation:

E = σ/ε = (F/A) / (ΔL/L)

E = (mg/(πd²/4) / (ΔL/L)

E = (4mg/(πd²) / (ΔL/L)

E = 4Lmg/(πd²ΔL)

E = 4(30.0)(90)(9.8)/(π(0.01²)0.25)

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2 years ago

What must be true about two objects if heat is flowing between them?

netineya [11]

Answer:

The objects must be different temperatures.

Explanation:

For heat to flow between two objects, heat must be flowing between them. The thermal gradient allows for the flow of heat. Heat is a form of energy that is dissipated from one place to another based on temperature difference.

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3 years ago

The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization

ANEK [815]

Answer:

The vapor pressure at 60.6°C is 330.89 mmHg

Explanation:

Applying Clausius Clapeyron Equation

$ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]$

Where;

P₂ is the final vapor pressure of benzene = ?

P₁ is the initial vapor pressure of benzene = 40.1 mmHg

T₂ is the final temperature of benzene = 60.6°C = 333.6 K

T₁ is the initial temperature of benzene = 7.6°C = 280.6 K

ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol

R is gas rate = 8.314 J/mol.k

$ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg$