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3 years ago

How does incident light that falls on an object affect the motion of electrons in the atoms of the object?

Physics

1 answer:

KonstantinChe [14]3 years ago

5 0

Explanation:

Incident light makes an electron oscillate. The electrons emit light or absorb the light, collide with other electrons, thereby converting light energy to more internal energy. and convert it to heat.

Light wave of a given frequency is incident on a material with electrons having the same vibrational frequencies, then electrons absorb the energy of the light wave and transform it into vibrational motion.

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What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +

Mumz [18]

Complete Question:

Given $\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$ at a point. What is the force per unit area at this point acting normal to the surface with $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$ ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, $\sigma_n = 28 MPa$

There are shear stresses acting on the surface since $\tau \neq 0$

Explanation:

$\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$

equation of the normal, $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$

$\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

Traction vector on n, $T_n = \sigma \b n$

$T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

$T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]$

$T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z$

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

$\sigma_n = T_n . \b n$

$\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa$

If the shear stress, $\tau$ , is calculated and it is not equal to zero, this means there are shear stresses.

$\tau = T_n - \sigma_n \b n$

$\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z$

$\tau = \sqrt{(-5/\sqrt{2})^2 + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa$

Since $\tau \neq 0$ , there are shear stresses acting on the surface.

3 0

3 years ago

PLEASE HELP ME!

PilotLPTM [1.2K]

Answer:

a) A = 3 cm, b) T = 0.4 s, f = 2.5 Hz,

2) A standing wave the displacement of the wave is canceled and only one oscillation remains

Explanation:

a) in an oscillatory movement the amplitude is the highest value of the signal in this case

A = 3 cm

b) the period of oscillation is the time it takes for the wave to repeat itself in this case

T = 0.4 s

the period is the inverse of the frequency

f = 1 /T

f = 1 /, 0.4

f = 2.5 Hz

2) a traveling wave is a wave for which as time increases the displacement increases, in the case of a transverse wave the oscillation is perpendicular to the displacement and in the case of a longitudinal wave the oscillation is in the same direction of the displacement.

A standing wave occurs when a traveling wave bounces off some object and there are two waves, one that travels in one direction and the other that travels in the opposite direction. In this case, the displacement of the wave is canceled and only one oscillation remains.

8 0

3 years ago

A car speeds up from 12.0 m/s to 16.0 m/s in 8.00s what is the acceleration

Lerok [7]

Answer:

0.5m/s²

Explanation:

acceleration =v-u/t

=(16-12)/8

=4/8

acceleration =0.5m/s²

7 0

3 years ago

Read 2 more answers

Jamie is at the state fair playing some games. At one booth he throws a 5-kg ball

Westkost [7]

Hey! How are you? My name is Maria, 19 years old. Yesterday broke up with a guy, looking for casual sex.

Write me here and I will give you my phone number - *pofsex.com*

My nickname - Lovely

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4 years ago

A house that was heated by electric resistance heaters consumed 1200 kWh of electric energy in a winter month. If this house wer

Marysya12 [62]

Answer:

$84

Explanation:

The coefficient of performance (COP) show the relationship between the power (kW) output of the heat pump and the power (kW) input to the compressor.

The heater consumed by the heater is 1200 kWh.

For a heat pump with a COP of 2.4, the electric input needed to produce an output of 1200 kWh is:

Electric input to heat pump = 1200 / 2.4 = 500 kWh

That means that supplying a heat pump with 500 kWh produces an output of 1200 kWh

The amount of power saved = power consumed by heater - power consumed by heat pump = 1200 - 500 = 700 kWh

Money saved = $0.12/kWh * 700 kWh = $84

6 0

3 years ago