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3 years ago

1. Which areas of physics deal with the following?

Physics

2 answers:

Genrish500 [490]3 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]3 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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What is the difference between kinetic and mechanical energy?

Wittaler [7]

Kinetic energy is energy that is in motion, thats all I remember

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3 years ago

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A small metal sphere has a mass of 0.14 g and a charge of -22.0 nc . it is 10 cm directly above an identical sphere with the sam

Allushta [10]

For this problem, we use the Coulomb's law written in equation as:

F = kQ₁Q₂/d²
where
F is the electrical force
k is a constant equal to 9×10⁹
Q₁ and Q₂ are the charge of the two objects
d is the distance between the two objects

Substituting the values:

F = (9×10⁹)(-22×10⁻⁹ C)(-22×10⁻⁹ C)/(0.10 m)²
F = 0.0004356 N

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3 years ago

A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit

Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c = 3 * 10⁸ Hz

Wavelength of the incident wave in air:

$\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m$

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

$n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3$

$\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m$

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

$n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

$n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }$

$n_1 = 377 \Omega$

The intrinsic impedance of media 2 is given as:

$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

Reflection coefficient, $r = \frac{n - n_{0} }{n + n_{0} }$

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is $E_{0} = 10 V/m$

Maximum amplitudes in the total field is given by:

$E = tE_{0}$ and $E = r E_{0}$

E = 10r, E = 10t

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3 years ago

The<br> is the particle in the nucleus with a positive charge.<br> Answer here

Finger [1]

Answer: A proton is a positively charged particle located in the nucleus of an atom.

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If a person weighs 600N on earth, how much does he weigh on the moon

Snezhnost [94]

Answer:

His weight would be 100 N

Explanation:

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