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vladimir1956 [14]
3 years ago
7

Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are

infinitely far away. What is the closest distance that she can see an object clearly when she wears her contacts?
Physics
1 answer:
Shalnov [3]3 years ago
4 0

Answer:

the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Explanation:

Given the data in the question,

near point = 20 cm

far point = 2 m = 200 cm

Now, for an object that is infinitely far away, the image is at is its far point.

so using the following expression, we can determine the focal length

1/f = 1/i + 1/o

where f is the focal length, i is the image distance and o is the object distance.

here, far point i = 2 m = 200 cm  and v is ∞

so we substitute

1/f = 1/(-200 cm)  +  1/∞

f = -200 cm

Also, for object at its closest point, the image appear at near point,

so

1/f = 1/i + 1/o

we make o the subject of formula

o = ( i × f ) / ( i - f )

given that near point i = 20 cm

we substitute

o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

o = 22.2 cm

Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

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babymother [125]

Answer:

E = 58.7 V/m

Explanation:

As we know that flux linked with the coil is given as

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here we have

A = \pi R_s^2

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now we have

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now the induced EMF is rate of change in magnetic flux

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now for induced electric field in the coil is linked with the EMF as

\int E. dL = EMF

E(2\pi r_c) = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L

E = \frac{\mu_o N^2 R_s^2 \frac{di}{dt}}{2 r_c L}

E = \frac{(4\pi \times 10^{-7})(6500^2)(0.14^2)(79)}{2(0.20)(3.50)}

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3 0
3 years ago
At 20°c, the resistance of a sample of nickel is 525 ohms. what is the resistance when the sample is heated to 70°C?​
Ipatiy [6.2K]

The resistance of the sample is 682.5\Omega

Explanation:

The relationship between resistance of a material and temperature is given by the equation

R(T)=R_0(1+\alpha (T-T_0))

where

R_0 is the resistance at the temperature T_0

\alpha is the temperature coefficient of resistance

For the sample of nickel in this problem, we have:

R_0 = 525 \Omega when the temperature is T_0 = 20^{\circ}C

While the temperature coefficient of resistance of nickel is

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Therefore, the resistance of the sample when its temperature is

T=70^{\circ}C

is

R=(525)(1+0.006(70-20))=682.5 \Omega

Learn more about resistance:

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Answer:

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Explanation:

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liraira [26]
F=ma
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Which object exerts more gravitational force, object A with a mass of 15 grams and a density of 2g/cm3 or object B with a mass o
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