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HACTEHA [7]
1 year ago
14

The area on an entrance ramp where you increase speed to that of expressway traffic is the ______? * 1 point deceleration lane m

edian lane entrance lane acceleration lane
Physics
1 answer:
lions [1.4K]1 year ago
8 0

The area on an entrance ramp where you increase speed to that of expressway traffic is the <u>acceleration lane.</u>

<u></u>

So the correct option is (D) that is acceleration lane.

A section or lane of adjustment for speed that has extra flooring on the borders of the lanes of traffic to let vehicles to accelerate until they merge with the flow of traffic.

Drivers must reach the posted speed limit before entering the acceleration lane, signal, locate a gap in traffic, and then merge.

Drivers can accelerate or decelerate in an area not being used by high-speed through traffic by using acceleration/deceleration lanes, also known as speed-change lanes or auxiliary lanes. The abrupt change in pace can result in stop-and-go traffic, crashes, and other problems on freeways and some major streets. These issues can be reduced by including acceleration/deceleration lanes in the roadway design.

Learn more about acceleration lane:

brainly.com/question/14344386

#SPJ4

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A simple electromagnet consisting of a coil of wire wrapped around an iron core. <u><em>A core of ferromagnetic material like iron serves to increase the magnetic field created.</em></u> The strength of magnetic field generated is proportional to the amount of current through the winding.

your answer is  b :)

I LOVE YOUR PROFILE PICTURE!!!

5 0
3 years ago
Do all protostars become stars?
Arada [10]
No not all of them. some can though.

7 0
3 years ago
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What is question 10—20 PTS!!
Galina-37 [17]
If my math is correct your answer should be 3.0 but correct me if I'm wrong
8 0
3 years ago
Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o
tankabanditka [31]

Answer:

(a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

Explanation:

Given that,

Power = 2.55 D

Object distance = 25 cm for near point

Object distance = ∞ for far point

Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D  is prescribed for distant vision?

(a) We need to calculate the focal length

Using formula of power

f =\dfrac{1}{P}

Put the value into the formula

f=\dfrac{100}{2.55}

f=39.21\ cm

We need to calculate the image distance

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}

\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}

\dfrac{1}{v}=-\dfrac{1421}{98025}

v=-68.98\ cm

The eye's near point is 68.98 cm from the eye.

(b). We need to calculate the focal length

Using formula of power

f =\dfrac{1}{P}

Put the value into the formula

f=-\dfrac{100}{3.00}

f=-33.33\ cm

We need to calculate the image distance

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}

-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}

\dfrac{1}{v}=-\dfrac{1}{33.33}

v=-33.33\ cm

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

5 0
3 years ago
In a physics experiment, a ball is released from rest, and it falls toward the ground. The timer was not paying attention but es
tigry1 [53]

Answer:

(A) –14m/s

(B) –42.0m

Explanation:

The complete solution can be found in the attachment below.

This involves the knowledge of motion under the action of gravity.

Check below for the full solution to the problem.

4 0
3 years ago
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