There are three things needed to be classified as a planet.
Pluto meets 2 of the 3.
The one it does not meet is it doesn’t “clear the neighborhood” which means a planet needs to be gravitationally dominant over all objects around it. Meaning smaller bodies ( asteroids) would either combine with the larger body making it bigger or pushing them away ( out of their orbit) so they do not collide with the larger body.
Answer:
q = 7.4 10⁻¹⁰ C
Explanation:
a) The magnetic force is given by the expression
F = q v x B
Where the blacks indicate vectors, q is the electric charge, v at particle velocity and B the magnitude of the magnetic field. If the velocity is perpendicular to the magnetic field, the sine is 1
F = q v B
Let's calculate the charge
q = F / vB
q = 1.00 10⁻¹² / 30.0 B
For the magnetic field of the earth we have a value between 25μT and 65μT, an intermediate value would be 45 μT, let's use this value.
q = 1 10⁻¹² / (30 45 10⁻⁶)
q = 7.4 10⁻¹⁰ C
b) In laboratories and modern electronics, currents of up to 1 10⁻⁶ A can be achieved without much difficulty, in advanced and research laboratories currents of up to 1 10⁻¹² can be managed. Load values (coulomb) cannot they are widely used today for work, but 1 mA = 3.6C, so we see that getting loads with the value of 10⁻¹⁰ C implies very small current less than 1 10⁻¹³ A, which only in laboratories of Very specialized can be created. Consequently, from the above it would be difficult to find loads lower than the calculated
The electrostatic charge is the one created by the friction between two surfaces, it is an indicated charge, in this case it would be possible to have better wing loads found from 10⁻¹⁰C
Answer:
7) λ = 0.5 m, 8) f = 4.8 10¹⁴ Hz
Explanation:
The speed of an electromagnetic wave is
c = λ f
where c is the speed of light in vacuum c = 3 10⁸ m / s
7) indicate the frequency f = 6.0 10⁸ Hz
we do not know the wavelength
λ = c / f
we calculate
λ = 3 10⁸ / 6.0 10⁸
λ = 0.5 m
8) indicate the wavelength λ = 6.25 10-7 m
we do not know the frequency
f = c / λ
we calculate
f = 3 10⁸ / 6.25 10⁻⁷
f = 0.48 10¹⁵ Hz
f = 4.8 10¹⁴ Hz
