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eimsori [14]
2 years ago
11

What's true about the elliptical path that the planets follow around the sun?

Physics
1 answer:
sattari [20]2 years ago
5 0

Answer:

well, it is not a solid path and thats a teeny tiny fact

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How many protons, neutrons, and electrons does a neutral atom of this element have? (round atomic mass to nearest whole number)
adelina 88 [10]

Answer:

The element "AI"  has:

Protons: 13 Neutrons: 14 Electrons: 13

Have a great day.

8 0
3 years ago
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Why do you think it is a driving force of weather around the world?
Finger [1]

Answer:

The sun.

Explanation:

The sun provides energy for living organisms, and it drives our planet’s weather and climate patterns.

Remember, Earth is spherical and the energy from the sun does not reach all areas with equal intensity. Areas exposed to the sun are directly on the sun’s rays (i.e. those nearest to the equator) and hence, receive greater solar input. In contrast, those in higher latitudes receive sunlight that is spread over a larger area and that has taken a longer path through the atmosphere. As a result, these higher latitudes receive less solar energy.

Also, ocean circulation and precipitation are all factors of weather

4 0
3 years ago
Define uniform and non uniform
cricket20 [7]

Answer:

When a body moves along a straight line with uniform speed or steady speed is called Uniform motion. When a body moves along a straight line but with variable or change in speed is called non-uniform motion.Hope this answer helps.

7 0
3 years ago
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If 100 kg person weighed 400 N on the planet Zorg,
Brut [27]

Answer:

<u>4 m/s²</u>

Explanation:

<u>Formula</u>

  • Weight = Mass x Gravity due to acceleration

<u>Solving</u> :

  • 400 N = 100 kg x g
  • g = 400/100
  • g = <u>4 m/s²</u>
5 0
2 years ago
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An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of
Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, E=390\ keV=390\times 10^3\ eV

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2meE}}

\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}

\lambda=1.96\times 10^{-12}\ m

(b) According to Bragg's law,

n\lambda=2d\ sin\theta

n = 1

For nickel, d=0.092\times 10^{-9}\ m

\theta=sin^{-1}(\dfrac{\lambda}{2d})

\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})

\theta=0.010^{\circ}

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0
3 years ago
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