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4 years ago

You should be extra careful during the hours of sunrise, sunset, and nighttime because

1 answer:

7 0

Poor visibility, difficulties in colour perception, lessened colour contrast vision due to darker shadows and reduced peripheral vision, these are the reasons why one has to be extra careful while driving during hours of sunrise, sunset and night time.

<h3><u> Explanation: </u></h3>

Sunrise, sunset and night time are parts of the day with minimal or absolutely no presence of sunlight. To safely navigate roads, we require enough light in order to detect presence of other vehicles, signs and pedestrians. Less sunlight during sunrise and sunset light the sky but makes the roads and vehicles have a darker, less bright view. The contrast between colours is the least, making it difficult to identify objects and see clearly.

A rising or a setting sun can also lead to glares on the driver’s view and thus obstruct it. Since a change in ambient light is observed, our eyes need to adjust with this change and this isn’t spontaneous. Night time driving has headlight glares from approaching vehicles and reduced surrounding visibility. The eyes switching for vision adaptability from dark to bright light if vehicles approach and pass by is not a quick action. Hence the driver’s vision is compromised in every such case and this may lead to accidents.

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How can I shorten this and for it to sill make sense?

Julli [10]

Answer:

Radiation is the emission or transmission of energy in the form of waves.

Explanation:

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3 years ago

Read 2 more answers

In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a

SOVA2 [1]

Answer: $0.084 m/s^2$

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance $d_1=8.13 km$

time taken=25 min =1500 s

initial speed $v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s$

after 8.13 km mark steve started to accelerate

speed after 60 s

$v_2=v_1+at$

$v_2=5.6+a\times 60$

distance traveled in 60 sec

$d_2=v_1\times 60+\frac{a60^2}{2}$

$d_2=336+1800 a$

time taken in last part of journey

$t_3=1663.6-1560=103.6 s$

distance traveled in this time

$d_3=v_2\times t_3$

$d_3=\left ( 5.6+a\times 60\right )103.6$

and total distance $=d_1+d_2+d_3$

$10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$a=0.084 m/s^2$

5 0

3 years ago

A(n) 93 kg clock initially at rest on a horizontal floor requires a(n) 610 N horizontal force to set it in motion. After the clo

denpristay [2]

Answer:0.669

Explanation:

Given

mass of clock 93 kg

Initial force required to move it 610 N

After clock sets in motion it requires a force of 514 N to keep moving it with a constant velocity

Initially static friction is acting which is more than kinetic friction

thus 613 force is required to overcome static friction

$\mu _smg=610$

$\mu _s\times 93\times 9.8=610$

$\mu _s=0.669$

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3 years ago

If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular

erastova [34]

Answer;

D. The car would begin to move in the direction it was headed in a straight line.

Explanation;

-Centripetal force is any net force causing uniform circular motion. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration.

-The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.

-Therefore,If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular path were reduced then the car would begin to move in the direction it was headed in a straight line.

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3 years ago

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Two objects experienced the same displacement in the same amount of time. The two objects must have the same -

nikdorinn [45]

If two object experiences the same displacement at the same time, the two objects must have the same average velocity.

The average velocity of an object is defined as the change in displacement per change in time of motion. This can be written as follows;

$avg. \ velocity = \frac{\Delta \ displacement }{\Delta \ time} \\\\V = \frac{\Delta \ x}{\Delta t}$

If two object experiences the same displacement at the same time, it means that the change in displacement with time is constant.

$\frac{\Delta x_1}{\Delta t_1} = \frac{\Delta x_2}{\Delta t_2} = V_{avg}$

Thus, we can conclude that if two object experiences the same displacement at the same time, the two objects must have the same average velocity.

Learn more here: brainly.com/question/23856383

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3 years ago