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3 years ago

9

At higher speeds, how would you compensate for the decrease in field of vision

2 answers:

inysia [295]3 years ago

5 0

When you ride a vehicle in a fast speed, then your peripheral vision will reduce that is why there is a need for you to follow the direction of the objects when you are travelling in order for you to compensate to the decrease in the field of vision.

gladu [14]3 years ago

5 0

<span>Your central and peripheral vision reduces whenever you ride a vehicle at a high speed. The best way to compensate for the decrease in your field of vision is to turn your head so that you can look to your sides. </span>

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3 years ago

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Zepler [3.9K]

Answer:

40mph

Explanation:

1st leg DATA:

time = 3 hrs ; speed = r mph ; distance = 3r miles

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2nd leg DATA:

speed = r mph ; distance = 12 miles

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2 years ago

What is the mass of an object that has a weight of 110N ?

fgiga [73]

Weight (W) = 110 N
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Let the mass of the object be m.
By using the formula, W = mg, we get,
110 N = 9.8 m/s^2 × m
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<u>Answer:</u>

<em><u>The </u></em><em><u>mass </u></em><em><u>of </u></em><em><u>the </u></em><em><u>object </u></em><em><u>is </u></em><em><u>1</u></em><em><u>1</u></em><em><u>.</u></em><em><u>2</u></em><em><u> </u></em><em><u>Kg.</u></em>

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2 years ago

A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the

Genrish500 [490]

Answer: Last option

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Explanation:

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$a_c =w^2r = \frac{v^2}{r}$

in this case we know the speed of the runner

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The radius "r" will be the distance from the runner to the center of the track

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$a_c = \frac{8^2}{28.2}\ m/s^2$

$a_c = 2.27\ m/s^2$

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3 years ago

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lara31 [8.8K]

Answer:

b

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