If one reverse the orientation of a permanent magnet ITS MAGNETIZATION WILL BE PERMANENTLY REVERSED. This is because, the magnetic domains inside the permanent magnet aligned with the new applied field and increase with it while those domains that are anti aligned with that field will shrink.
Answer:
i think you minus The weights I think
Answer:
c. Solar eclipses would be much more frequent.
Explanation:
The <u>ecliptic plane</u> is the apparent orbit that the sun describes around the earth (although it is the earth that orbits the sun), is the path the sun follows in earth's sky.
A <u>solar eclipse</u> occurs when the moon gets between the earth and the sun, so a shadow is cast on the earth because the light from the sun is blocked.
The reason why solar eclipses are not very frequent is because the moon's orbital plane is not in the same plane as the orbit of the earth around the sun, but rather that it is somewhat inclined with respect to it.
So <u>if both orbits were aligned, the moon would interpose between the sun and the earth more frequently, producing more solar eclipses.</u>
So, if the moon's orbital plane were exacly the same as the ecliptic plane solar eclipses would be more frequent.
the answer is: c.
Answer:
The answer is A and C.
Explanation:
Only two factors are relevant when dealing with the gravitational force between two objects - their mass and their distance apart from one another. Gravity's force is proportional to the product of the masses of the two objects and is inversely proportional to the square of the distance between them.
Answer:
1,85 m / s²
Explanation:
De la pregunta anterior, se obtuvieron los siguientes datos:
Velocidad inicial (u) = 40 km / h
Hora inicial (t₁) = 0
Tiempo final (t₂) = 6 s
Velocidad final (v) = 0
Aceleración (a) =?
A continuación, convertiremos 40 km / ha m / s. Esto se puede obtener de la siguiente manera:
1 km / h = 0,2778 m / s
Por lo tanto,
40 km / h = 40 km / h × 0,2778 m / s / 1 km / h
40 km / h = 11,11 m / s
Por tanto, 40 km / h equivalen a 11,11 m / s.
Finalmente, determinaremos la aceleración del móvil durante el período en el que desaceleró. Esto se puede obtener de la siguiente manera:
Velocidad inicial (u) = 11,11 m / s
Hora inicial (t₁) = 0
Tiempo final (t₂) = 6 s
Velocidad final (v) = 0
Aceleración (a) =?
a = (v - u) / (t₂ - t₁)
a = (0 - 11,11) / (6 - 0)
a = - 11,11 / 6
a = –1,85 m / s²
Por tanto, la aceleración del móvil durante el período en el que se ralentizó es de –1,85 m / s²
