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16 . You are turning onto a two-lane road divided by a broken yellow line. You know immediately that:

Over [174]

When a person is turning onto a two-lane road divided by a broken yellow line, you know immediately that you are on a two-way road.

<h3>What is the road about?</h3>

Note that a Yellow centerlines can be seen in roads and it is one that is often used to separate traffic moving in different directions.

Note also that Broken lines can be crossed to allow slower-moving traffic and as such, When a person is turning onto a two-lane road divided by a broken yellow line, you know immediately that you are on a two-way road.

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You are turning onto a two-lane road divided by a broken yellow line. You know immediately that:

Answers

You are on a two-way road.

You are on a one-way road.

The road is under repair.

You must stay to the left of the broken yellow lines.

Learn more about two-way road from

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5 0

2 years ago

What type of engineer makes sure equipment is safe and operational

zvonat [6]

Answer:

mechanical engineer is the best answer

8 0

3 years ago

What should you, the worker, be aware of with regard to evacuation procedures at your workplace

Alinara [238K]

Answer:

As a worker, it is important to follow the proper set of instructions or emergency plans during an emergent situation. Not carefully following the rules may result to a bigger problem such as further injury and damage to property.

Explanation:

Evacuation Procedure- This is a step-by-step procedure that people follow in order to safely vacate any building or place. This procedure is applicable to any situation, such as the workplace. This is now called the <em>Workplace Evacuation Procedure. </em>This is very important because there are so many unpredictable situations or events that are happening in the world right now, such as fire or earthquake. This procedure is being done through an evacuation plan.

The awareness of the workers regarding the proper way to evacuate during emergency situation is very important. It will be easier for them to know where to locate the nearest exit route. They will also learn to stop any form of device or equipment that could cause a hazzard during the situation. In case of the hospital, which is also a workplace, the employees will also learn how to assist the patients before themselves. They will also know where to assemble if there's a need to do so.

7 0

4 years ago

Read 2 more answers

Determine the nature of the following cycle (reversible, irreversible, or impossible): a refrigeration cycle draws heat from a c

vlabodo [156]

Answer:

Impossible.

Explanation:

The ideal Coefficient of Performance is:

$COP_{i} = \frac{250\,K}{300\,K-250\,K}$

$COP_{i} = 5$

The real Coefficient of Performance is:

$COP_{r} = \frac{950\,kJ-70\,kJ}{70\,kJ}$

$COP_{r} = 12.571$

Which leads to an absurds, since the real Coefficient of Performance must be equal to or lesser than ideal Coefficient of Performance. Then, the cycle is impossible, since it violates the Second Law of Thermodynamics.

6 0

3 years ago

Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

viktelen [127]

Answer:

$\theta_1=15^o\\\theta_2=75^o$

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

$V_x=V_{ox}=V_ocos\theta$

Where $\theta$ is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

$V_y=V_{oy}-gt=V_osin\theta-gt$

The horizontal and vertical distances are, respectively:

$x=V_{o}cos\theta t$

$\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}$

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

$\displaystyle t_f=\frac{2V_osin\theta}{g}$

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

$\displaystyle R=\frac{V_o^2sin2\theta}{g}$

Let's solve for $\theta$

$\displaystyle sin2\theta=\frac{R.g}{V_o^2}$

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

$\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}$