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4 years ago

What is a common way to correct nearsightedness?

Physics

2 answers:

Jlenok [28]4 years ago

8 0

Answer: Place a concave lens in front of the eye.

Explanation:

Hi, a common way to to correct nearsightedness is to Place a concave lens in front of the eye.

These types of lenses are usually used in eyeglasses that correct nearsightedness.

The problem that nearsighted people have is that the distance between the eye's lens and retina is longer than it should be. It makes impossible to distinguish distant objects.

Placing concave lenses in front of a nearsighted eye reduces the refraction of light and lengthens the focal length so that the image is formed on the retina.

xxMikexx [17]4 years ago

4 0

Concave lens , they are thinnest at the center and thicker at the edge

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A gas at 5.00 atm pressure was stored in a tank during the winter at 5.0 °C. During the summer, the temperature in the storage a

saw5 [17]

Answer:

a) 5.63 atm

Explanation:

We can use combined gas law

<em>The combined gas law</em> combines the three gas laws:

Boyle's Law, (P₁V₁ =P₂V₂)
Charles' Law (V₁/T₁ =V₂/T₂)
Gay-Lussac's Law. (P₁/T₁ =P₂/T₂)

It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.

P₁V₁/T₁ =P₂V₂/T₂

where P = Pressure, T = Absolute temperature, V = Volume occupied

The volume of the system remains constant,

So, P₁/T₁ =P₂/T₂

a) $\frac{5}{278} =\frac{P_2}{313} \\\\P_2=\frac{5*313}{278}\\ P_2 = 5.63 atm$

7 0

4 years ago

An electromagnetic wave is a kind of a wave which,

Harman [31]

Explanation:

Electromagnetic waves are waves that propagate through space while an electric field and a magnetic field interact with each other.

please mark as brainliest plzz

7 0

3 years ago

Read 2 more answers

A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg

alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = - $\frac{v}{u}$

where

u = object distance

v = image distance

Now,

1.90 = $\frac{v}{u}$

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

$\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}$

$\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}$

$\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}$

$\frac{1}{16.5} = \frac{ 0.90}{1.90u}$

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0

4 years ago

A skier of mass 60 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him 75 m up a 30° slop

postnew [5]

Answer:

Explanation:

Given

mass of skier=60 kg

distance traveled by skier=75 m

inclination $(\theta )=30^{\circ}$

speed (v)=2.4 m/s

as the skier is moving up with a constant velocity therefore net force is zero

$F_{net}=0$

Force applied by cable $=mg\sin \theta$

$F=60\times 9.8\times \sin (30)=294 N$

work done $=F\cdot x$

$W=294\cdot 75=22.125 J$

(b)Power $=F\cdot v$

$P=294\cdot 2.4=705.6 W\approx 0.946 hp$

5 0

3 years ago

What is the energy range (in joules) of photons of wavelength 410 nm to 750 nm ? Express your answers using two significant figu

andreyandreev [35.5K]

Answer:

4.9 x 10^-19 J, 2.7 x 10^-19 J

Explanation:

first wavelength, λ1 = 410 nm = 410 x 10^-9 m

Second wavelength, λ2 = 750 nm = 750 x 10^-9 m

The relation between the energy and the wavelength is given by

E = h c / λ

Where, h is the Plank's constant and c be the velocity of light.

h = 6.63 x 10^-34 Js

c = 3 x 10^8 m/s

So, energy correspond to first wavelength

E1 = (6.63 x 10^-34 x 3 x 10^8) / (410 x 10^-9) = 4.85 x 10^-19 J

E1 = 4.9 x 10^-19 J

So, energy correspond to second wavelength

E2 = (6.63 x 10^-34 x 3 x 10^8) / (750 x 10^-9) = 2.652 x 10^-19 J

E2 = 2.7 x 10^-19 J

4 0

3 years ago