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Marina86 [1]
3 years ago
13

what did classical physics predict would happen to the light given of by an object as its tempurtare increased

Physics
2 answers:
alexgriva [62]3 years ago
7 0
When the temperature of an object that is giving off light is increased, the particles in the object will move at a faster rate and there will be increased vibration of these molecules. This will makes the object to emit more light and to shine more brightly.
VikaD [51]3 years ago
6 0

Answer:

The energy of the light would increase from visible light into the ultraviolet range.

Explanation:

blah blah blah, blah blah BLAH

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Which of the following describes the charge of an atom before any electrons are transferred?
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The answer is neutral charge. An atom element will always and has to be stable, in order for this state to happen. The charge of an electron has to be neutral. For atom with neutral charge, the proton will always equal the number of electron.

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A boy runs at a speed of 3.3 m/s straight off the end of a diving board that is 3 meters above the water. How long is he airborn
svp [43]

Answer:

45

Explanation:

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Read this /https://www.carbonbrief.org/polar-bears-and-climate-change-what-does-the-science-say
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Melting ice would damage this polar bears habitat meaning the polar bear may decrease


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3 years ago
Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib
Alexus [3.1K]

Answer:

<em>a) 318.2 W/m^2</em>

<em>b) 2.5 x 10^-4 J</em>

<em>c) 1.55 x 10^-8 v/m</em>

<em></em>

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = \pi d^{2} /4 = (3.142 x (2*10^{-3} )^{2})/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>

<em></em>

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>

<em></em>

c) The peak electric field is given as

E = \sqrt{2I/ce_{0} }

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

e_{0} = 8.85 x 10^9 m kg s^-2 A^-2

E = \sqrt{2*318.2/3*10^8*8.85*10^9}  = <em>1.55 x 10^-8 v/m</em>

6 0
4 years ago
A wind turbine is rotating counterclockwise at 0.626 rev/s and slows to a stop in 12.9 s. Its blades are 17.9 m in length. What
iogann1982 [59]

Answer:

276.5 m/s^2

Explanation:

The initial angular velocity of the turbine is

\omega=0.626 rev/s \cdot 2\pi rad/rev =3.93 rad/s

The length of the blade is

r = 17.9 m

So the centripetal acceleration is given by

a=\omega^2 r

At the instant t = 0,

\omega=3.93 rad/s

So the centripetal acceleration of the tip of the blades is

a=(3.93 rad/s)^2 (17.9 m)=276.5 m/s^2

5 0
3 years ago
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