Answer: See below
Explanation:
The Earth attracts the falling object with the same intensity of gravity as the object attracts the Earth, according to Newton's law of gravitation. The displacement of the two bodies, however, is inversely proportional to their respective masses.
Example: The Earth attracts a ball that falls 3 metres from the ground, even though the ball's mass is insignificant in comparison to the Earth's. Similarly, the ball draws the Earth with the same power, but the Earth's mass is enormously more than the ball's. As a result, the Earth collides with a billionth of a millimetre ball (or even less). Restart the Earth's descent on the ball you'll never see again.
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Answer:
Explanation:
We know that,
Neptune is 4.5×10^9 km from the sun
And given that,
Earth is 1.5×10^8km from sun
Then,
Let P be the orbital period and
Let a be the semi-major axis
Using Keplers third law
Then, the relation between the orbital period and the semi major axis is
P² ∝ a³
Then,
P² = ka³
P²/a³ = k
So,
P(earth)²/a(earth)³ = P(neptune)² / a(neptune)³
Period of earth P(earth) =1year
Semi major axis of earth is
a(earth) = 1.5×10^8km
The semi major axis of Neptune is
a (Neptune) = 4.5×10^9km
So,
P(E)²/a(E)³ = P(N)² / a(N)³
1² / (1.5×10^8)³ = P(N)² / (4.5×10^9)³
Cross multiply
P(N)² = (4.5×10^9)³ / (1.5×10^8)³
P(N)² = 27000
P(N) =√27000
P(N) = 164.32years
The period of Neptune is 164.32years
Answer:
Tiempo, t = 6.11 segundos.
Explanation:
Dados los siguientes datos;
Distancia = 140 m
Aceleración = 7,5 m/s²
Dado que el objeto partió del reposo, su velocidad inicial es igual a 0 m/s.
Para encontrar el tiempo, usaríamos la segunda ecuación de movimiento;
S = ut + ½at²
Dónde;
S representa el desplazamiento o la altura medida en metros.
u representa la velocidad inicial medida en metros por segundo.
t representa el tiempo medido en segundos.
a representa la aceleración medida en metros por segundo cuadrado.
Sustituyendo en la ecuación, tenemos;
140 = 0*t + ½*7.5*t²
140 = 0 + 3.75t²
140 = 3.75t²
Dividiendo ambos lados por 3,75, tenemos;
t² = 140/3.75
t = √37.33
Tiempo, t = 6.11 segundos
I think the amplitude changed
