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Umnica [9.8K]
3 years ago
9

lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this l

ens
Physics
1 answer:
ser-zykov [4K]3 years ago
4 0
<h2>Question: </h2>

The nearpoint of an eye is 151 cm. A corrective lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this lens?

Answer:

29.96cm

Explanation:

Using the corrective lens, the image should be formed at the front of the eye and be upright and virtual.

Now using the lens equation as follows;

\frac{1}{f} = \frac{1}{v} + \frac{1}{u}   -------------(i)

Where;

f = focal length of the lens

v = image distance as seen by the lens

u = object distance from the lens

From the question;

v = -151cm        [-ve since the image formed is virtual]

u = 25cm

Rewrite equation (i) to have;

f = \frac{uv}{u+v}

Substitute the values of v and u into the equation;

f = \frac{25*(-151)}{25-151}

f = \frac{-3775}{-126}

f = 29.96cm

The focal length should be 29.96cm

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  • Initial velocity

Kinematics is the part of physics that establishes the relationships between the position, velocity, and acceleration of bodies.

In this case we have a vertical launch

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Where y and y₀ are the final and initial positions, respectively, v₀ the initial velocity, g the acceleration of gravity (g = 9.8 m / s²) and t the time

   

With the ball in hand, its position is zero

         0 = 0 + v₀ t - ½ g t²

         v₀ t - ½ g t² = 0

         v₀ = ½ g t

Let's calculate

         v₀ = ½ 9.8 2.4

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3 years ago
Four charges with equal magnitudes of 10.6 × 10-12 C are placed at the corners of a rectangle. The lengths of the sides of the r
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Answer:

Figure a. E_net = 99.518 N/C

Figure b. E_net = 177.151 N / C

Explanation:

Given:

- Attachment for figures missing in the question.

- The dimensions for rectangle are = 7.79 x 3.99 cm

- All four charges have equal magnitude Q = 10.6*10^-12 C

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Find the magnitude of the electric field at the center of the rectangle in Figures a and b.

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                                         E = k*Q / r^2

- Where, k is the coulomb's constant = 8.99 * 10^9

Part a)

- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:

                                 E_1 + E_3 = 0

- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:

                                  E_net = E_2 + E_4

                                  E_2 = E_4

                                  E_net = 2*E = 2*k*Q / r^2

- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:

                                  r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

                                  r = sqrt ( 1.9151*10^-3 ) = 0.043762 m

- Plug the values in the E_net expression developed above:

                                  E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3

                                 E_net = 99.518 N/C

Part b)

- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).

- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.

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                                  Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.

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                                  E_net = 177.151 N / C

               

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