Physics, 26.01.2020 07:31 greatsavagebeast

12 3 4 5 6 7 8 9 10 time remaining 59: 49 which changes would result in a decrease in the gravitational force between two objects? check all that apply.

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Answer from: carlo123

increasing the distance between the objects

decreasing the mass of one of the objects

decreasing the mass of both objects

Explanation:

The complete question would be:

Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.

increasing the distance between the objects

decreasing the distance between the objects

increasing the mass of one of the objects

increasing the mass of both objects

decreasing the mass of one of the objects

decreasing the mass of both objects

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Gravitational force between two objects can be computed using the formula:

$Fg = G\dfrac{Mm}{r^2}$

Where:

Fg is the gravitational force (Newtons or N)

G is the gravitational constant (6.674⋅10⁻¹¹N(m/kg)²)

M is the mass of one object (Kilograms or Kg)

m is the mass of the other object (Kilograms or Kg)

r is the distance between the two objects (m)

Now notice the relationship between the mass and gravitational force. Their relationship is direct, meaning as the value of one goes up, the other goes up and the reverse holds true. If the mass of one or both of the objects decreases, the gravitational force also decreases.

Now look at the relationship between the distance between the objects and the gravitational force-- they are indirect. This means that as one increases, the other decreases. So if we increase the distance between the objects, the gravitational force between them decreases as well.

Answer from: adhanom12

increasing the distance between the objects

decreasing the mass of one of the objects

decreasing the mass of both objects

Explanation:

Answer from: Quest

What do you need on

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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