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Sophie [7]
2 years ago
8

What should a driver do when he/she gets to a speed hump? a maintain current speed. b accelerate. c brake. d reduce speed to sug

gested limit.
Physics
1 answer:
Llana [10]2 years ago
6 0

It is usually best to reduce speed to suggested limit when  a driver gets to a speed hump and is denoted as option D.

<h3>What is Deceleration?</h3>

This refers to a situation in which the speed or acceleration of a body is reduced while in the other hand, acceleration refers to the condition in which a body increases its speed or acceleration.

Speed humps are structures which are placed in the road in other to slow vehicles to between 10 to 15 mph. They ensure over speeding is checked due to it being dangerous to other users.

Applying brake or maintaining current speed may lead to the vehicle being tilted towards the front end which may result in collision and other forms of accident.

This therefore the reason why reducing speed to suggested limit is the most appropriate choice.

Read more about Speed hump here brainly.com/question/9213903

#SPJ1

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3 years ago
Bill is farsighted and has a near point located 121 cm from his eyes. Anne is also farsighted, but her near point is 74.0 cm fro
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Answer:

Explanation:

The lens equation is

1 / f = 1 / di + 1 / do

Where

f is focal length

di is the image distance

do is the object distance

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Then, the object distance is

do = 25 - 2 = 23cm

The have the same object distance.

Let find the vocal length of bills eye

Given that,

Bill near point is 121cm and distance of the glass from the eye is 2cm

Then,

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object distance do = 23cm

Then,

1 / f_B = 1 / di_B + 1 / do

1 / f_B = -1 / 119 + 1 / 23

1 / f_B = -119^-1 + 23^-1

1 / f_B = 0.0351

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Also, let find Annie focal length

Given that,

Annie near point is 74 cm and distance of the glass from the eye is 2cm

Then,

Image distance of Annie is

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object distance do = 23cm

Then,

1 / f_A = 1 / di_A + 1 / do

1 / f_A = -1 / 72 + 1 / 23

1 / f_A = -72^-1 + 23^-1

1 / f_A = 0.02959

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Distance of object from the lens when Annie uses Billy glass

Then,

1 / f_B = 1 / di_A + 1 / do

1 / 28.51 = -1 / 72 + 1 / do

28.51^-1 = -72^-1 + do^-1

do^-1 = 28.51^-1 + 72^-1

do^-1 = 0.048964

do = 20.42 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_A = 20.42 + 2 = 22.42cm

do_A = 22.42 cm

Distance of object from the lens when Billy uses Annie glass

Then,

1 / f_A = 1 / di_B + 1 / do

1 / 33.8 = -1 / 119 + 1 / do

33.8^-1 = -119^-1 + do^-1

do^-1 = 33.8^-1 + 119^-1

do^-1 = 0.03799

do = 26.32 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_B = 26.32 + 2 = 28.32 cm

do_B = 28.32 cm

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