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Paha777 [63]
3 years ago
6

After a check up, a person now has a far point of 100 cm, but with good near point vision. He needs to wear a new pair of correc

tive glasses. He accidentally takes his old contact lenses with him while traveling. If the lenses of the old pair have a power of -0.5 diopters, what is his far point (measured from his eye) when he is wearing the old lenses?
Physics
1 answer:
lakkis [162]3 years ago
7 0

Answer:

so his far point according to this pair of glass is 200 cm

Explanation:

power of old pair of corrective glasses is given as

P = -0.5 dioptre

now we have

f = \frac{1}{P}

f = -2 m

f = -200 cm

now we know that for normal vision the maximum distance of vision is for infinite distance

so by lens formula we have

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

\frac{1}{d_i} + 0 = \frac{1}{200}

d_i = 200 cm

so his far point according to this pair of glass is 200 cm

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Hola, para resolver debemos convertir unidades utilizando equivalencias

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The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ
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Explanation:

At point B, the velocity speed of the train is as follows.

          \nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})

                           = (30)^{2} + 2(-0.25(412 - 0))

                           = 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

          y = 200 e^{\frac{x}{1000}}

      \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}

Now, second derivative of the train is calculated as follows.

         \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}      

       \frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}    

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            a_{n} = \frac{\nu^{2}_{B}}{\rho}

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Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is 0.309 m/s^{2}.

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