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Dahasolnce [82]
3 years ago
6

An individual is nearsighted; his near point is 13.0 cm and his far point is 50.4 cm. what lens power is needed to correct his n

earsightedness?
Physics
1 answer:
Colt1911 [192]3 years ago
5 0

Answer:

= -1.984 Diopters

Explanation:

The lens should form an upright, virtual image at the far point.

Therefore;

The image distance, V will be -50.4 cm, since the image is virtual.

For objects at very far distant the object distance,u, will be ∞ ; This means that focal length, f, will be equivalent to image distance, v, that is -50.4 cm

Therefore; f = -50.4/100 = -0.504 m

But, since Power of a lens, P, is given by the reciprocal of focal length in meters, (1/f)

Then, power will be given by;

Power = 1/f

           = 1/-0.504 m

           =- 1.984

Power is measured in Diopters

Hence          <u>= -1.984 Diopters</u>

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Sveta_85 [38]

Answer:

(2) −1 e

Explanation:

A quark is the lightest elementary particles which form hadron such as proton and neutron. A quark has fractional charge.

Up, charm and top quarks have +\frac{2}{3} e charge where as down, strange and bottom quarks have -\frac{1}{3}e charge.

The antiparticle of up quark is antiup quark and has charge -\frac{2}{3}e charge.

The antiparticle of down quark is antidown quark and has charge +\frac{1}{3}e charge.

An antibaryon is composed of two anti-up quark and one anti-down quark.

Net charge of the anti-baryon is:

2\times (-\frac{2}{3} e)+1\times (+\frac{1}{3})e=-1e

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5 0
4 years ago
What type of circuit is seen in this picture?<br><br> -Open<br> -Parallel<br> -Series<br> -AC
MA_775_DIABLO [31]

it's a. open circuit

3 0
3 years ago
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sineoko [7]
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7 0
3 years ago
The potential difference between two parallel conducting plates in vacuum is 165 V. An alpha particle with mass of 6.50×10-27 kg
shepuryov [24]

Answer:

kinetic energy (K.E) = 5.28 ×10⁻¹⁷            

Explanation:

Given:

Mass of  α particle (m) = 6.50 × 10⁻²⁷ kg

Charge of  α particle (q) = 3.20 × 10⁻¹⁹ C

Potential difference ΔV = 165 V

Find:

kinetic energy (K.E)

Computation:

kinetic energy (K.E) = (ΔV)(q)

kinetic energy (K.E) = (165)(3.20×10⁻¹⁹)

kinetic energy (K.E) = 528 (10⁻¹⁹)

kinetic energy (K.E) = 5.28 ×10⁻¹⁷              

4 0
3 years ago
A silver bar 0.125 meter long is subjected to a temperature change from 200°C to 100°C. What will be the length of the bar after
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\Delta L=  \alpha L_0 (T_f-T_i)

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248

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3 years ago
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