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3 years ago

When describing image formation in mirrors, what is the angle of incidence?

2 answers:

5 0

'B' is the correct choice.

BUT ... the angle of incidence is not the angle between the light ray
and the mirror. It's the angle between the light ray and the NORMAL
to the mirror. The 'normal' is the line that's perpendicular to the mirror.

Eva8 [605]3 years ago

5 0

Answer:

B. The angle the light ray makes as it strikes the surface.

Explanation:

Angle of the incidence is the angle that incident light ray will make with the normal to the surface at which light will strike. The surface is reflecting surface and here it follow the law of reflection

As per law of reflection of light angle of incidence with respect to the normal must be equal to the angle of reflection with normal in other direction.

So we can say that

angle of incidence = angle of reflection

so here we can say that the correct answer would be

B. The angle the light ray makes as it strikes the surface.

so the light ray will reflect at the same angle as that of angle of incidence given in it

so it must be

$\theta_i = \theta_r$

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Convert 13.1 miles to feet. Using one step conversion

Marat540 [252]

Answer:

69,168 ft

Explanation:

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3 years ago

Light travels through a liquid at 2.25e8 m/s. What is the index of refraction of the liquid?

Pavlova-9 [17]

Answer:

The index of refraction of the liquid is n = 1.33 equivalent to that of water

Explanation:

Solution:-

- The index of refraction of light in a medium ( n ) determines the degree of "bending" of light in that medium.

- The index of refraction is material property and proportional to density of the material.

- The denser the material the slower the light will move through associated with considerable diffraction angles.

- The lighter the material the faster the light pass through the material without being diffracted as much.

- So, in the other words index of refraction can be expressed as how fast or slow light passes through a medium.

- The reference of comparison of how fast or slow the light is the value of c = 3.0*10^8 m/s i.e speed of light in vacuum or also assumed to be the case for air.

- so we can mathematically express the index of refraction as a ratio of light speed in the material specified and speed of light.

- The light passes through a liquid with speed v = 2.25*10^8 m/s :

$n = c / v\\\\n = \frac{ 3*10^8 }{2.25*10^8} \\\\n = 1.33$

- The index of refraction of the liquid is n = 1.33 equivalent to that of water.

8 0

3 years ago

Read 2 more answers

The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard

gogolik [260]

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

$|\Delta E_{ij}| = |E_i-E_j|$

Our states are given by

$E_1 = -11.7eV$

$E_2 = -4.2eV$

$E_3 = -3.3eV$

In this way the energy balance for the states would be given by,

$|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\$

$|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV$

$|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV$

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

8 0

3 years ago

A man runs at a velocity of 6.2 m/s for 11.5 minutes. When going up and increasingly steep hill, he slows down at a constant rat

Tanzania [10]

6.2 times 11.5 the divide 0.25

3 0

3 years ago

Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of

marusya05 [52]

Answer:

a) $\dot m = 6.878\,\frac{kg}{s}$ , b) $T = 104.3^{\textdegree}C$ , c) $\dot S_{gen} = 11.8\,\frac{kW}{K}$

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

$-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0$

The mass flow rate is:

$\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}$

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

$P = 7000\,kPa$

$T = 500^{\textdegree}C$

$h = 3411.4\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

State 2s - Liquid-Vapor Mixture

$P = 100\,kPa$

$h = 2467.32\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

$x = 0.908$

The isentropic efficiency is given by the following expression:

$\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}$

The real specific enthalpy at outlet is:

$h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})$

$h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )$

$h_{2} = 2684.46\,\frac{kJ}{kg}$

State 2 - Superheated Vapor

$P = 100\,kPa$

$T = 104.3^{\textdegree}C$

$h = 2684.46\,\frac{kJ}{kg}$

$s = 7.3829\,\frac{kJ}{kg\cdot K}$

The mass flow rate is:

$\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}$

$\dot m = 6.878\,\frac{kg}{s}$

b) The temperature at the turbine exit is:

$T = 104.3^{\textdegree}C$

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

$\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0$

$\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})$

$\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )$

$\dot S_{gen} = 11.8\,\frac{kW}{K}$

4 0

3 years ago