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Novay_Z [31]
4 years ago
8

An object is placed 45 cm in front of a crystal ball. The crystal ball has a radius of 14 cm, and is made from a glass with an i

ndex of refraction of n = 1.5.
Find the position of the final image once the rays have gone through the two faces, as measured from the second face. Is a virtual object formed anywhere in the analysis? Yes or no
Physics
1 answer:
abruzzese [7]4 years ago
6 0

Answer:

Explanation:

 Using the formula for refraction at curved surface

For refraction at first surface

μ₂ /v -μ₁ /u = ( μ₂ - μ₁ )/R

1.5/v -1/-45 =.5/14

v = 111 cm

For refraction at opposite surface ,

u ( object distance ) = 111 - 28 =83 cm

It will act as virtual object for surface 2 so it will be positive in sign

again using the above equation for 2 nd surface we have

1/v - 1.5/83.17 = -.5-=/-14 = 1/28

v = 18.62 cm from second face.

yes, virtual image will be formed.

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A boulder of mass 2000. kg breaks loose of a mountainside and plunges 200. m straight down into a lake at a temperature of 5.00◦
natka813 [3]

Answer:

DS = 13865.7[J/K]

Explanation:

We can calculate the energy of the rock, like the potential energy relative to the lake level. Which can be calculated by means of the following expression of the potential energy:

E_{p}=m*g*h\\\\where:\\m = mass = 2000[kg]\\h = elevation = 200 [m]\\g = gravity = 9.81[m/s^2]

Therefore:

E_{p}=2000*9.81*200\\E_{p}=3924000 [J]\\

This energy is transformed into thermal energy.

we shall remember that isothermal heat transfer processes are internally reversible, so the entropy change of a system during one of these processes can be determined, by the following expression.

DS=\frac{Q}{T}\\ where:\\DS = entropy change [J/K]\\Q = Heat transfer [J]\\T = temperature [K]

T = 5 + 278 = 283[K]

DS = 3924000 / 283

DS = 13865.7[J/K]

5 0
3 years ago
You send a beam of light from a material with index of refraction 1.21 into an unknown material. In order to help identify this
White raven [17]

Answer:

refractive index of the unknown material is 1.33.

Explanation:

μ₁ = 1.21

incidence angle (i) = 41.9°

refraction angle (r) = 37.3°

Let us assume μ be the refractive index of the unknown material

according to snell's law of refraction.

μ₁ sin i = μ₂ sin r

1.21 × sin 41.9° =  μ × sin 37.3°

μ = 1.33

hence the refractive index of the unknown material comes out top be 1.33

6 0
3 years ago
Indica qué es una propiedad específica de la materia. Además explica por qué son útiles las propiedades específicas de la materi
Katyanochek1 [597]

Answer:

Check Explanation

Comprobar explicación

Explanation:

English Translation

Indicate what a specific property of matter is. Also explain why the specific properties of matter are useful compared to the general ones.

Solution

The specific properties of matter are properties that describes the intensive properties of the system. They are properties that do not depend on or change with the extent or size of the system. They are usually obtained by dividing the generalised properties or extensive properties by the extent or size of matter to make them independent of size/extent/Mass.

Examples of specific properties include specific heat capacity, specific volume etc. They usually have units of general units/Mass units.

The specific properties of matter are more important than the general ones because

- They help in general comparisons of the properties of different materials. They are used to rank, classify and compare properties of different materials.

- They are used in reference table/data to easily record easily accessible properties of matter. It helps to record standards that are general and independent of sizes/extents/Mass, thereby keeping the reference table/data/chart precise and concise.

- They provide us with values that are easy to memorize and remember, unlike trying to cram the different properties of different masses/sizes of matter.

In Spanish/En español

Las propiedades específicas de la materia son propiedades que describen las propiedades intensivas del sistema. Son propiedades que no dependen ni cambian con la extensión o el tamaño del sistema. Por lo general, se obtienen dividiendo las propiedades generalizadas o las propiedades extensivas por la extensión o el tamaño de la materia para hacerlas independientes del tamaño / extensión / masa.

Los ejemplos de propiedades específicas incluyen capacidad calorífica específica, volumen específico, etc. Usualmente tienen unidades de unidades generales / unidades de masa.

Las propiedades específicas de la materia son más importantes que las generales porque

- Ayudan en las comparaciones generales de las propiedades de diferentes materiales. Se utilizan para clasificar, clasificar y comparar propiedades de diferentes materiales.

- Se utilizan en la tabla / datos de referencia para registrar fácilmente propiedades de materia fácilmente accesibles. Ayuda a registrar estándares que son generales e independientes de tamaños / extensiones / masa, manteniendo así la tabla / datos / tabla de referencia precisa y concisa.

- Nos proporcionan valores que son fáciles de memorizar y recordar, a diferencia de tratar de agrupar las diferentes propiedades de diferentes masas / tamaños de materia.

Hope this Helps!!!

¡¡¡Espero que esto ayude!!!

7 0
3 years ago
A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains
Colt1911 [192]

If only I was smart then I could help you :/ no but like for real im madddd dumb sorry :(

4 0
4 years ago
Energy transferred as heat occurs between two bodies in thermal contract when they differ in which of the following properties?
Hatshy [7]
Temperature is the answer
8 0
3 years ago
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