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serg [7]
3 years ago
9

Which Astronomer estimated the relative size and distance of the moon and sun?

Physics
1 answer:
N76 [4]3 years ago
8 0
Eratosthenes is the astronomer
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A student in your class whom you kind-of like asks you come to watch a meteor shower. What exactly are you being invited to? a.
madam [21]

Answer:D. Watching the left-over dirt from a comet burn up by friction as the pieces hit the Earth's atmosphere

Explanation: A meteor shower is giant rocks moving across the sky

5 0
2 years ago
Write the minkowski version of newtons seconds law in terms of proper acceleration
Sindrei [870]
F=ma where f=force,m=mass,acceleration =a
4 0
3 years ago
The mass of a proton is approximately equal to
Dahasolnce [82]

Answer:

1.6726\cdot 10^{-27} kg

Explanation:

The three main particles that make an atom are:

- Proton: its mass is 1.6726\cdot 10^{-27} kg, it carries an electric charge of +e (e=1.6\cdot 10^{-19}C), and it is located in the nucles of the atom

- Neutron: its mass is 1.6749 \cdot 10^{-27}kg, it carries no electric charge, and it is also located in the nucleus of the atom

- Electron: its mass is 9.1094 \cdot 10^{-31}kg, it carries an electric charge of -e (e=1.6\cdot 10^{-19}C), and it is located outside the nucleus

3 0
3 years ago
Which statement is true about the inner planets of our solar system
poizon [28]
B they formed from the dense elements as compared to the others in the solar system
4 0
3 years ago
The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 ​5 ​​ m/s then ho
Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10​⁵​​ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

F = \frac{MV}{t}

solving this two equations together;

\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9}  \ s \ = 1.639 \ ns

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0
3 years ago
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