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quester [9]
3 years ago
11

A student analyzes data of the motion of a planet as it orbits a star that is in deep space. The orbit of the planet is consider

ed to be stable and does not change over time. What can the student say about this initial observation?
Physics
2 answers:
Valentin [98]3 years ago
8 0

Explanation:

The orbit of the planet is considered to be stable and does not change over time. This means that the force of gravity and inertia are perfectly balanced. Leading to forward motion of Planet and moons under force of gravity of some other large body(Most probably stars).This is in turns leads to formation of orbit.

Naily [24]3 years ago
4 0

Answer:

The student is correct. Since the planet move around the stars orbit in a stable manner, the only force that can put it in space and make it move round the star will be the force of gravity from the star.

Explanation:

A student analyzes data of the motion of a planet as it orbits a star that is in deep space. The orbit of the planet is considered to be stable and does not change over time. What can the student say about this initial observation?

The student is correct. Since the planet move around the stars orbit in a stable manner, the only force that can put it in space and make it move round the star will be the force of gravity from the star.

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By what potential difference must a proton [m_0 = 1.67E-27 kg) be accelerated to have a wavelength lambda = 4.23E-12 m? By what
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Explanation:

1. Mass of the proton, m_p=1.67\times 10^{-27}\ kg

Wavelength, \lambda_p=4.23\times 10^{-12}\ m

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

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

V=\dfrac{h^2}{2q_pm_p\lambda^2}

V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 1.67\times 10^{-27}\times (4.23\times 10^{-12})^2}

V = 45.83 volts

2. Mass of the electron, m_p=9.1\times 10^{-31}\ kg

Wavelength, \lambda_p=4.23\times 10^{-12}\ m

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

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

V=\dfrac{h^2}{2q_em_e\lambda^2}

V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times (4.23\times 10^{-12})^2}

V=6.92\times 10^{34}\ V

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Hence, this is the required solution.

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