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4 years ago

1. Which areas of physics deal with the following?

Physics

2 answers:

Genrish500 [490]4 years ago

8 0

Answer:

1. how fast things move - mechanics

2. how the shape of a cave affects an echo - Sound & acoustics

3. which sunglasses are best for cutting the glare on a ski slope - optics

4. how the cooling system in a refrigerator works - thermodynamics

5. what lightning is - Electricity

6. how energy is produced by the sun – nuclear physics

Explanation:

Mechanics - the study of the motions of the objects

Sound - the study of the behavior of sound waves

Optics - the study of the behavior of light waves and optical instruments

Thermodynamics - the study of the relationship between heat and other forms of energy

Electricity - the study of properties of charges, their movement and their field interactions

Nuclear physics - study of atomic nucleus and its properties

lana66690 [7]4 years ago

5 0

ANSWER:

a. Kinematics.

b. Waves/Acoustics.

c. Optics/Polarization.

d. Thermal/Conservation of Energy.

e. Electricity and magnetism.

f. Renewable energy

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2 years ago

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Leno4ka [110]

Answer: $1.593*10^{17} photons$ released if the laser is used 0.056 s during the surgery

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First, you have to calculate the energy of each photon according to Einstein's theoty, given by:

$E =\frac{hc}{\lambda}$

Where $\lambda$ is the wavelength, $h$ is the Planck's constant and $h$ is the speed of light

$h = 6.626*10^{-34} \frac{m^{2} kg }{s}$ -> Planck's constant

$c = 3*10^{8} \frac{m}{s}$ -> Speed of light

So, replacing in the equation:

$E =\frac{ 6.626*10^{-34} \frac{m^{2} kg }{s}*3*10^{8} \frac{m}{s}}{514*10^-9 m}$

Then, the energy of each released photon by the laser is:

$E = 3.867*10^{-19} \frac{J}{photons}$

After, you do the inverse of the energy per phothon and as a result, you will have the number of photons in a Joule of energy:

$\frac{1}{3.867*10^{-19}} = 2.586*10^{18} \frac{photons}{J}$

The power of the laser is 1.1 W, or 1.1 J/s, that means that you can calculate how many photons the laser realease every second:

$2.586*10^{18}\frac{photons}{J} * 1.1 \frac{J}{s} = 2.844*10^{18} \frac{photons}{s}$

And by doing a simple rule of three, if $2.844*10^{18} photons$ are released every second, then in 0.056 s:

$0.056 s*2.844*10^{18} \frac{photons}{s} = 1.593*10^{17} photons$ are released during the surgery

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3 years ago