Determina la fuerza de acción que se presenta entre dos cuerpo cuyas masas son 500 Kg y 900 Kg, si ambos cuerpos se encuentran s
1 answer:
Answer:
F = 7.50 10⁻⁶ N
Explanation:
The force of attraction between two bodies is given by the law of universal gravitation
F =
where m₁ and m₂ are the masses of the hides, r the distance that separates them and G the constant of attraction that is worth 6.67 10⁻¹¹ N m²/kg²
let's calculate
F = 6.67 10⁻¹¹ 500 900/2²
F = 7.50 10⁻⁶ N
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Answer:
a) p=0, b) p=0, c) p= ∞
Explanation:
In quantum mechanics the moment operator is given by
p = - i h’ d φ / dx
h’= h / 2π
We apply this equation to the given wave functions
a) φ =
.d φ dx = i k
We replace
p = h’ k
i i = -1
The exponential is a sine and cosine function, so its measured value is zero, so the average moment is zero
p = 0
b) φ = cos kx
p = h’ k sen kx
The average sine function is zero,
p = 0
c) φ =
d φ / dx = -a 2x
.p = i a g ’2x
The average moment is
p = (p₂ + p₁) / 2
p = i a h ’(-∞ + ∞)
p = ∞
Answer:
The final position made with the vertical is 2.77 m.
Explanation:
Given;
initial velocity of the ball, V = 17 m/s
angle of projection, θ = 30⁰
time of motion, t = 1.3 s
The vertical component of the velocity is calculated as;
The final position made with the vertical (Yf) after 1.3 seconds is calculated as;
Therefore, the final position made with the vertical is 2.77 m.