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Nataliya [291]
3 years ago
13

A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision

Physics
1 answer:
BigorU [14]3 years ago
7 0

Complete Question

The  complete question is  shown on the first uploaded image  

Answer:

a

  f=  -75 \ cm =  - 0.75 \ m

b

  P  =  -1.33 \ diopters

Explanation:

From the question we are told that

    The  image distance is  d_i =  -75 cm

The value of the image is negative because it is on the same side with the corrective glasses

    The  object distance is  d_o =  \infty

The  reason object distance  is because the object father than it being picture by the eye

General focal length is mathematically represented as

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

substituting values

             \frac{1}{f}  =  \frac{1}{-75}  -   \frac{1}{\infty}

=>         f=  -75 \ cm =  - 0.75 \ m

Generally the power of the corrective lens is  mathematically represented as

        P  =  \frac{1}{f}

substituting values

       P  =  \frac{1}{-0.75}

        P  =  -1.33 \ diopters

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A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2000 kg SUV traveling from ea
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v = 11.1m/s

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2 years ago
A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie
Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, V_{Hall} = 20\ mV = 0.02\ V

Magnetic Field, B = 0.10 T

Hall emf, V'_{Hall} = 69\ mV = 0.069\ V

Now,

Drift velocity, v_{d} = \frac{V_{Hall}}{B}

v_{d} = \frac{0.02}{0.10} = 0.2\ m/s

Now, the expression for the electric field is given by:

E_{Hall} = Bv_{d}sin\theta                            (1)

And

E_{Hall} = V_{Hall}d

Thus eqn (1) becomes

V_{Hall}d = dBv_{d}sin\theta

where

d = distance

B = \frac{V_{Hall}}{v_{d}sin\theta}                      (2)

(a) When \theta = 90^{\circ}

B = \frac{0.069}{0.2\times sin90} = 0.345\ T

(b) When \theta = 60^{\circ}

B = \frac{0.069}{0.2\times sin60} = 0.398\ T

5 0
2 years ago
a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigg
IrinaK [193]

Answer:

f = 1 m

Explanation:

The magnification of the lens is given by the formula:

M = \frac{q}{p}

where,

M = Magnification = 4

q = image distance = 5 m

p = object distance = ?

Therefore,

4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m

Now using thin lens formula:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\

<u>f = 1 m</u>

6 0
2 years ago
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Fiesta28 [93]

Answer:

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7 0
2 years ago
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