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In which direction does a convergent boundary move?

VladimirAG [237]

Answer:

Toward each other teehee merry christmas

Explanation:

7 0

3 years ago

Read 2 more answers

Who is responsible of refraction light 1/3 in eye

alexandr1967 [171]

The cornea is responsible of refraction light 1/3 in eye.

<h3>What is the function of the cornea?</h3>

In addition to the protective function, it plays a fundamental role in the formation of vision. Transparent, it works like a lens over the iris, focusing light from the pupil towards the retina.

Normally, the cornea and lens deflect (refract) incoming light rays, focusing them on the retina. The shape of the cornea is fixed, but the lens changes shape to focus on objects at different distances from the eye.

See more about cornea at brainly.com/question/2297282

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3 0

2 years ago

Find the wavelength of the balmer series spectral line corresponding to n = 15.

Anna71 [15]

Answer:

371.2 mm

Explanation:

The Balmer series of spectral lines is obtained from the formula

1/λ = R(1/2² -1/n²) where λ = wavelength, R = Rydberg's constant = 1.097 × 10⁷ m⁻¹

when n = 15

1/λ = 1.097 × 10⁷ m⁻¹(1/2² -1/15²)

= 1.097 × 10⁷ m⁻¹(1/4 -1/225)

= 1.097 × 10⁷ m⁻¹(0.25 - 0.0044)

= 1.097 × 10⁷ m⁻¹ 0.245556

= 2.693 10⁶ m⁻¹

So,

λ = 1/2.693 10⁶ m⁻¹

= 0.3712 10⁻⁶ m

= 371.2 mm

7 0

3 years ago

A 49 N coconut is in a 15 m tree. What is te potential energy of the coconut?

eduard

Answer:

735 J

Explanation:

From the question given above, the following data were obtained:

Weight (W) = 49 N

Height (h) = 15 m

Potential energy =?

Potential energy is simply defined as the product of weight of the object and height to which the object is raised. Mathematically, it is expressed as:

Potential energy = weight × height

With the above formula, we can obtain the potential energy of the coconut as follow:

Weight (W) = 49 N

Height (h) = 15 m

Potential energy =?

Potential energy = weight × height

Potential energy = 49 × 15

Potential energy = 735 J

Thus, the potential energy of the coconut is 735 J

4 0

3 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

4 years ago