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Elena-2011 [213]
3 years ago
15

the gravitational force between two objects is 1600 and what will be the gravitational force between the objects if the distance

between them doubles?
Physics
1 answer:
antoniya [11.8K]3 years ago
7 0
The equation for gravitational force is:

F = GMm/r^2

Where G is a constant, M is the mass of the larger object, m is the mass of the other object, and r is the distance between the two objects.

So:

1600 = GMm/r^2

Now we can find the force when the distance doubles:

F = GMm/(2r)^2

GMm will remain the same since G is a constant and the masses didn't change, so the only difference between this and the first calculation is the denominator.

By simplifying the denominator, we find:

F = GMm/(4r^2)

Now we can see that the only difference between this and the equation given the original distance is that we are dividing by 4. This means that the force will be one fourth of the original force:

1600/4 = 400

The answer is 400.
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Se golpea una pelota de golf de manera que su velocidad inicial forma un ángulo de 45° con la horizontal. La pelota alcanza el s
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Answer:

42m/s

6.06s

Explanation:

To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:

x_{max}=\frac{v_o^2sin(2\theta)}{g}\\\\x_{max}=vt_{max}

θ: angle = 45°

vo: initial velocity

g: gravitational constant = 9.8m/s^2

x_max: max distance = 180 m

t_max: max time

by replacing the values of the parameters and do vo the subject of the first formula you obtain:

v_o=\sqrt{\frac{gx_{max}}{sin(2\theta)}}\\\\v_o=\sqrt{\frac{(9.8m/s^2)(180m)}{sin(2(45\°))}}=42\frac{m}{s}

with this value of vo you calculate the max time:

t_{max}=\frac{x_{max}}{v}=\frac{x_{max}}{v_ocos(45\°)}\\\\t_{max}=\frac{180m}{(42m/s)cos(45\°)}=6.06s

hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s

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TRANSLATION:

Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:

θ: ángulo = 45 °

vo: velocidad inicial

g: constante gravitacional = 9.8m / s ^ 2

x_max: distancia máxima = 180 m

t_max: tiempo máximo

reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:

con este valor de vo usted calcula el tiempo máximo:

por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s

4 0
3 years ago
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