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Irina-Kira [14]
2 years ago
14

When a projecting load extends to the rear four or more feet beyond the bed or body of a vehicle in the daytime, the extreme rea

r and sides of the load must be marked by four?
Physics
2 answers:
ella [17]2 years ago
7 0
Four red flags

The load being carried poses a hazard to vehicles in the proximity of the load-carrying vehicle, and so it is mandatory by law for the vehicle to make the load more visible. At night time, the load is marked using red lights, as flags would not be visible then.
DedPeter [7]2 years ago
4 0

Four red flags as markers.

<h2>Further Explanation </h2>

The cargo that is carried poses a danger to the surrounding vehicles. vehicles carrying more loads are required to use this mark because it is an obligation that the loaded vehicles are more visible. At night, the load is marked using a red light, because the flag will not be seen at that time. It is also mandatory for the driver to carry/install a safety triangle.

A safety triangle is a triangle-shaped sign with a red side that is used to secure a place of accident or blow. Traffic regulations in various countries require that all vehicles have safety triangles and use them when vehicles must be stopped in dangerous places.

Safety triangles sold today are usually made of plastic and can be folded. If the vehicle is forced to be stopped because of an accident, strike, broken tire, etc., the safety triangle is placed in the direction of traffic, usually behind the car. The red color as intended, must be able to reflect light when hit by light, and finally, its position must cross the road with a pointed angle facing up, and the red color facing the direction of traffic. The security triangle must be colored,

The red color as intended, must be able to reflect light when struck by the light beam, and finally, its position must cross the road with a pointed angle facing up, and the red color facing the direction of traffic.

Learn more

driving safety brainly.com/question/4306793

Details

Grade:  High School

Subject:  Physics

keywords:  driving safety

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Explanation:

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3 years ago
What are non-contact forces please include example in your answer
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3 years ago
A car takes off from rest takes of from rest and covers a distance of 80m on a straight road in 10s.Calculate the magnitude of i
hodyreva [135]
  • Initial velocity (u) = 0 m/s [the car was at rest]
  • Distance (s) = 80 m
  • Time (t) = 10 s
  • Let the magnitude of acceleration be a.
  • By using the equation of motion, s = ut +  \frac{1}{2} a {t}^{2}we get,80 = 0 \times 10 +  \frac{1}{2}  \times a \times  {10}^{2}  \\  =  > 80 =  \frac{1}{2}  \times 100a \\  =  > 80 = 50a \\  =  > a =  \frac{80}{50}  \\  =  > a = 1.6

<u>A</u><u>nswer:</u>

<u>The </u><u>magnitude</u><u> </u><u>of </u><u>its </u><u>acceleration</u><u> </u><u>is </u><u>1</u><u>.</u><u>6</u><u> </u><u>m/</u><u>s^</u><u>2</u><u>.</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

4 0
3 years ago
A ball is thrown up at a speed of 20 m/s.
ioda

Given:

(Initial velocity)u=20 m/s

At the maximum height the final velocity of the ball is 0.

Also since it is a free falling object the acceleration acting on the ball is due to gravity g.

Thus a=- 9.8 m/s^2

Now consider the equation

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Where v is the final velocity which is measured in m/s

Where u is the initial velocity which is measured in m/s

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s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.

Substituting the given values in the above formula we get

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s= 400/19.6= 20.41m

Thus the maximum height attained is 20.41 m by the ball

6 0
3 years ago
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