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devlian [24]
3 years ago
8

PLEASE HELP!!!!!!

Physics
1 answer:
SashulF [63]3 years ago
7 0


Somewhere in your book or your notes, you must have met the formula for the
gravitational attraction between two bodies.  If you can go back and find it, you
only need to plug your numbers into that formula, and out will pop the answer.

Formula:                            <u>Force = G (mA x mB) / (distance)²</u>

If everything is in SI units, then        G = 6.67 x10⁻¹¹ newton-meter² / kilogram²

You said that
                         mA = 8.1 kg
                         mB = 6.5 kg
                 distance = 0.5 m .

   Force = (6.67 x 10⁻¹¹ nt-m²/kg²) (8.1kg x 6.5kg / (0.5m)² =

                (6.67 x 10⁻¹¹ nt-m²/kg²)    ( 52.65 kg² ) / (0.25 m²) =

                     <em>1.4047 x 10⁻⁸ newtons .</em>

That's roughly    5.052 x 10⁻⁸ ounce .  (5% of one micro-ounce)


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The study of motion is called kinematics.

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Answer:

a. El tiempo de recorrido es 5.882\times 10^{-3} segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

b. El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

Explanation:

El sonido es un tipo de onda mecánica, que es un tipo de onda que necesita de un medio material para propagarse. En este caso, entendemos que el sonido se propaga a través del aire atmosférico hasta llegar a su destino y devolverse a rapidez constante. Entonces, podemos estimar el tiempo (t), medido en segundos, a partir de la siguiente fórmula:

t = \frac{2\cdot x_{s}}{v_{s}}

Donde:

x_{s} - Distancia entre la cámara fotográfica y el objeto, medida en metros.

v_{s} - Rapidez del sonido en el aire atmosférico, medida en metros por segundo.

A continuación, calculamos el tiempo de recorrido:

a. (x_{s} = 1\,m, v_{s} = 340\,\frac{m}{s})

t = \frac{2\cdot (1\,m)}{340\,\frac{m}{s} }

t = 5.882\times 10^{-3}\,s

El tiempo de recorrido es 5.882\times 10^{-3} segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

b. (x_{s} = 20\,m, v_{s} = 340\,\frac{m}{s})

t = \frac{2\cdot (20\,m)}{340\,\frac{m}{s} }

t = 0.118\,s

El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

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