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goblinko [34]
3 years ago
14

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest

2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness
Physics
1 answer:
shutvik [7]3 years ago
5 0

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

Answer:  3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

gap between lense and eye = 2cm

Therefore;

image distance d₁ = - 45cm = - 0.45m

object distance  d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f =  5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

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Can inserting a resistor in a circuit produce an effect similar to a short circuit?
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2 years ago
An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. Aft
sergeinik [125]

Answer:

a) 80 V

b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

Explanation:

<u>Given :</u>

We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( K_{A}= 0) to the point B at distance l = 0.500m where its kinetic energy is (  K_{B}= 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = + 30.0 v.

<u>Required :</u>

<em>(a) We are asked to find the electric potential VB </em>

<em>(b) We want to determine the magnitude and the direction of the electric field E. </em>

<u> Solution </u>

(a) We are given the values for VA,K_{B} and q so we want to find a relationship between these three parameters and VB to get the value of VB.

As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:

                                   K_{A} +U_{A} =K_{B} +U_{B} .........................................(1)                                          

Where K_{A}= 0 and the potential energy U of the charge is given by U = q V

where V is the electric potential.  So, equation (1) will be in the form :

                                  0+qVA=K_{B} +qVB                      (Divide by q)

                                         VA=K_{B} /q + VB                  (solve for VB)

                                         VB=VA- K_{B}/q .......................................(2)

We get the relation between VB, VA and K_{B}, now we can plug our values for VA, K_{B} and q into equation (2) to get VB

                                         VB=VA- K_{B}/q

                                              =30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)

                                              =80 V

(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points

                                   VA-VB =\int\limits^1_0 {E} \, dl...................................(a)

                                               =E\int\limits^1_0 {} \, dl

                                   VA-VB=El                      (solve for E)

                                            E= VA-VB/l..................................(3)

Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E

                                            E= VA-VB/l

                                              =-100 N/C

The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

5 0
3 years ago
A 80 ohms resistor, 0.2 H inductance and 0.1 mF capacitor are connected in series across a generator (60 Hz, V rms=120 V). Deter
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Answer:

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Current = 1.81 A

Explanation:

Resistance = R = 80 ohms

Inductance = L = 0.2 H

Inductive reactance = XL = X_{L}= = ωL = (2πf) L

= 2 (3.14) (60)(0.2) = 75.398 Ohms

Capacitive reactance = 1 / ωC = 1/(2πf)C = 1 / [(2π)(60)(0.1 × 10⁻3)]

= 26.526 Ohms  

Impedance = Z = \sqrt{R^{2} + (X_{L} - X_{C})^{^{2}}} =

= \sqrt{8788.511} = 93.747 ohms  

Voltage = \sqrt{2} × 120 = 169.7056 V

Current = I = V ÷ R = (169.7056) ÷ 93,747 = 1.81 A

8 0
3 years ago
If 34.7 g of O2 reacts with iron to form 79.34 g of iron oxide, how much iron was used in the reaction?
zhuklara [117]

Answer: B. 44.64 g

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Mass of reactants = mass of iron + mass of oxygen = mass of iron + 34.7 g

Mass of product = mass of iron oxide = 79.34 g

As Mass of reactants = Mass of product

mass of iron + 34.7 g = 79.34 g

mass of iron = 44.64 g

Thus 44.64 g of iron was used in the reaction

6 0
3 years ago
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