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2 years ago

15

Calculate the focal length of a lens needed by a woman whose near point is 50cm from her eyes, assuming the least distance of di

stinct vision for a normal eye is 25cm.

1 answer:

son4ous [18]2 years ago

4 0

The focal length of a lens needed by a woman whose near point is 50cm from her eyes is 50cm.

To find the answer, we have to know about the focal length of correcting lens.

<h3>How to find the focal length of correcting lens?</h3>

If x is the distance of nearest point of the defective eye and D is the least distance of distinct vision, then, the expression for focal length of the correcting lens will be,

$f=\frac{XD}{X-D}$

It is given that, the woman whose near point is 50cm from her eyes, assuming the least distance of distinct vision for a normal eye is 25cm. Thus, the focal length will be,

$f=\frac{50*25}{50-25} =50cm.$

Thus, we can conclude that, the focal length of a lens needed by a woman whose near point is 50cm from her eyes is 50cm.

Learn more about the focal length here:

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However, sometimes levers are designed to increase the input force needed to move an object. Which of the following is most like

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3 years ago

An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average mag

Afina-wow [57]

Answer:

The induced emf in the loop is $7.35\times 10^{-4}\ V$

Explanation:

Given that,

Length of the wire, L = 1.22 m

It changes its shape is changed from square to circular. Then the side of square be its circumference, 4a = L

4a = 1.22

a = 0.305 m

Area of square, $A=a^2=(0.305)^2=0.0930\ m^2$

Circumference of the loop,

$C=2\pi r=L\\\\r=\dfrac{L}{2\pi}\\\\r=\dfrac{1.22}{2\pi}=0.194\ m$

Area of circle,

$A'=\pi r^2\\A'=\pi (0.194)^2\\\\A'=0.118\ m^2$

The induced emf is given by :

$\epsilon=\dfrac{\d\phi}{dt}\\\\\epsilon=\dfrac{\d(BA)}{dt}\\\\\epsilon=B\dfrac{A'-A}{t}\\\\\epsilon=0.125 \times \dfrac{0.118-0.0930}{4.25}\\\\\epsilon=7.35\times 10^{-4}\ V$

So, the induced emf in the loop is $7.35\times 10^{-4}\ V$

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3 years ago

If you started the motor of a boat in the middle of a lake, who would detect the sound of the motor first: a friend sitting on t

ArbitrLikvidat [17]

Answer:

A friend snorkeling just below the surface of the water along the same shore will detect the sound first.

Explanation:

The speed of sound in water medium is faster than that through the air.
Sound propagates through the medium by transferring through the molecules on it. Water has more closely packed molecules due to which the speed is faster.
In fact, the sound's speed in water is almost four times faster than that in the air.
So the guy in the water surface gets to hear sound faster than the one in sore.

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4 years ago

Read 2 more answers

A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t

enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

$v_{avg} = \frac{6000}{20*60}$

$v_{avg} = 5m/s$

now since effect of wind is in opposite direction then we can say

$V_{net} = v_{bird} - v_{wind}$

$5 = 9 - v_{wind}$

$v_{wind}= 4 m/s$

Part b)

now if bird travels in the same direction of wind then we will have

$v_{net}= v_{bird} + v_{wind}$

$v_{net} = 9 + 4 = 13 m/s$

now we can find the time to go back

$time = \frac{distance}{speed}$

$time = \frac{6000}{13}$

$time = 7.7 minutes$

Part c)

Total time of round trip when wind is present

$T = t_1 + t_2$

$T = 20 + 7.7 = 27.7 min$

now when there is no wind total time is given by

$T = \frac{6000}{9} + \frac{6000}{9}$

$T = 22.22 min$

So due to wind time will be more

4 0

4 years ago

A stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. What is the spe

lana [24]

Answer: V = 15 m/s

Explanation:

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F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz

Using doppler effect formula

F = C/ ( C - V) × f

Where

F = observed frequency

f = source frequency

C = speed of light = 3×10^8

V = speed of the car

Substitute all the parameters into the formula

2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10

2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)

1.000000049 = 3×10^8/(3×10^8 - V)

Cross multiply

300000014.7 - 1.000000049V = 3×10^8

Collect the like terms

1.000000049V = 14.71429

Make V the subject of formula

V = 14.71429/1.000000049

V = 14.7 m/s

The speed of the car is 15 m/s approximately

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3 years ago