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Mamont248 [21]
3 years ago
6

Complete the ray diagram below:

Physics
2 answers:
-BARSIC- [3]3 years ago
4 0

Answer:

<em>C. Inverted, real, and larger</em>

Explanation:

Inverted - An inverted image is formed when the rays intersect below the principal axis. An inverted image points downward.

Real - A real image can be projected on a screen. A real image is on the opposite side of lens as compared to the object. Whereas, for a mirror, a real image forms on the same side as the object.

Image size - a general measure of the image as compared to the original object. Image size is typically stated as smaller than the object, the same size as the object, or larger than the object.

liubo4ka [24]3 years ago
3 0

Answer:

inverted, real, and larger

Explanation:

To form the image of the object placed at the 5cm mark, the ray diagram must be properly drawn as described.

1. The first incident ray coming from the object strike the lens and parallel to the principal axis will converge at the other side of the lens and pass through the focus after refraction.

2. The second incident ray must pass through the optical center of the lens undeflected.

The refracted ray in (1) and the incident ray in (2) will intersect at a point where the image will then be formed.

The nature of the image that wikk be formed will be inverted, real and larger than the object as shown in the attachment.

Note that inverted images are real images.

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A ball is rolled at a velocity of 25 m/sec. after 13.5 seconds, it comes to a stop . What is the acceleration of the ball?
nexus9112 [7]

Answer:

The answer to your question is a = -1.85 m/s² the acceleration is negative because it is coming to stop.

Explanation:

Data

vo = 25 m/s

t = 13.5 s

a= ?

vf = 0 m/s

Formula

vf = vo + at

solve for a

a = (vf - vo)/t

Substitution

a = (0 - 25) / 13.5

Simplification

a = -25/13.5

Result

a = -1.85 m/s²

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7 0
3 years ago
A point charge is placed at each corner of square with side leanth a. The charges all have same magnitude q. My question, What i
nexus9112 [7]

Answer:

 E = k q / a²   (1.3535) (- i ^ + j ^)

  E = k q / a²  1.914  ,      θ’= 135

Explanation:

For this exercise we will use Newton's second law where we must add as vectors

        E_total = E₁₂ i ^ + E₁₄ j ^ + E₁₃

Let's look for the value of each term

On the x axis

       E₁₂ = k q / a²

On the y axis

       E₁₄ = k q / a²

For the charge in the opposite corner we look for the distance

        d = √ (a² + a²) = a √2

let's look for the field

      E₁₃ = k q / d²

      E₁₃ = k q / 2a²

let's use trigonometry to find the two components of this field

       cos 45 = E₁₃ₓ / E₁₃

       E₁₃ₓ = E₁₃ cos 45

       

       sin 45 = E_{13y} / E₁₃

       E_{13y} = E₁₃ sin 45

       E₁₃ₓ = k q / 2a²  cos 45

       E_{13y} = k q / 2a²  sin 45

let's find each component of the electric field

X axis

      Eₓ = -E₁₂ - E₁₃ₓ

      Eₓ = - k q / a² - k q / 2a² cos 45

      Eₓ = - k q / a² (1 + cos 45/2)

      cos 45 = sin 45 = 0.707

      Eₓ = - k q / a²   (1 + 0.707 / 2)

      Eₓ = - k q / a²    (1.3535)

Y axis  

      E_y = E₁₄ + E_{13y}

       E_y = k q / a² + k q / 2a²     sin 45

       E_y = k q / a² (1 + sin 45/2)

       E_y = k q / a²       (1.3535)

we can give the results in two ways

       E = k q / a²   (1.3535) (- i ^ + j ^)

In modulus and angle form, let's use Pythagoras' theorem for the angle

       E = √ (Eₓ² + E_y²)

        E = k q / a²    1.3535 √2

        E = k q / a²     1.914

we use trigonometry for the angle

        tan θ = E_y / Eₓ

         θ = tan⁻¹  (E_y / Eₓ)

         θ = tan⁻¹ (1 / -1)

         θ = 45

in the third quadrant, if we measure the angle of the positive side of the x-axis

           θ‘= 90 + 45

           θ’= 135

4 0
3 years ago
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