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3 years ago

A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision.

Physics

1 answer:

Pavel [41]3 years ago

7 0

Answer:

(a) nearsighted

(b) diverging

(c) the lens strength in diopters is 1.33 D, and considering the convention for divergent lenses normally prescribed as: -1 33 D

Explanation:

(a) The person is nearsighted because he/she cannot see objects at distances larger than 75 cm.

(b) the type of correcting lens has to be such that it counteracts the excessive converging power of the eye of the person, so the lens has to be diverging (which by the way carries by convention a negative focal length)

$f=\frac{1}{d} =\frac{1}{0.75} = 1.33\,D$

So it would normally be written with a negative signs in front indicating a divergent lens.

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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors

Mars2501 [29]

Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

Given :

Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

The length of the capacitor $(l)$ = $3 \times 10^{-3}$ m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

capacitance per unit length = $\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}$

(B)

The charge on both conductors is given by,

$Q = C \Delta V$

Where, C = capacitance of cylindrical capacitor and value of $C = 12.06 \times 10^{-13}$ F, $\Delta V = 350 \times 10^{-3}$ V

∴ $Q = 4.22 \times 10^{-19}$ C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is $+Q$ and Charge on outer conductor is $-Q$ .

8 0

3 years ago

Republicans are called the Liberal Party while the Democrats are called Conservative Party.

Fudgin [204]

Answer:

Explanation:

False

5 0

3 years ago

Read 2 more answers

Please follow me guys <br>

Lilit [14]

Answer:

if i do give me brainliest ok ok

Explanation:

7 0

3 years ago

A 77.0−kg short-track ice skater is racing at a speed of 12.6 m/s when he falls down and slides across the ice into a padded wal

sukhopar [10]

Answer:

-6112.26 J

Explanation:

The initial kinetic energy, $KE_i$ is given by

$KE_i=0.5mv_1^{2$ } where m is the mass of a body and $v_i$ is the initial velocity

The final kinetic energy, $KE_f$ is given by

$KE_f=0.5mv_f^{2}$ where $v_f$ is the final velocity

Change in kinetic energy, $\triangle KE$ is given by

$\triangle KE=KE_f-KE_i=0.5mv_f^{2}-0.5mv_1^{2}=0.5m(v_f^{2}-v_i^{2})$

Since the skater finally comes to rest, the final velocity is zero. Substituting 0 for $v_f$ and 12.6 m/s for $v_i$ and 77 Kg for m we obtain

$\triangle KE=0.5*77*0^{2}-0.5*77*(0^{2}-12.6^{2})=-6112.26 J$

From work energy theorem, work done by a force is equal to the change in kinetic energy hence for this case work done equals <u>-6112.26 J</u>

3 0

3 years ago

If Karen rode her bicycle 5 km to school in 30 minutes, what was her average speed? Give your answers in km/h.

Doss [256]

Answer:

10 km/h

Explanation:

Before calculating the average speed, we need to convert the time taken for the trip from minutes to hours. Since 1 hour cointains 60 minutes, we have:

$1 h: 60 min = t : 30 min\\t=\frac{1 h \cdot 30 min}{60 min}=0.5 h$

Karen's average speed is given by:

$v=\frac{d}{t}$

where

d = 5 km is the distance covered

t = 0.5 h is the time taken for the trip

Substituting into the equation, we find

$v=\frac{5 km}{0.5 h}=10 km/h$

5 0

3 years ago