Physics, 21.08.2019 01:30, manco6

Angular momentum of a particle is conserved when f = 0.

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answered: Learsyguerra26

True

Explanation:

As we know that the torque can be written as,

$\tau=r\times F$

According to the question total forece, F=0

This total null force will make the torque zero.

And torque is related to angular momentum as,

$\tau=\frac{dL}{dt}$

Here, L is angular momentum.

Now,

$\frac{dL}{dt}-=0\\ L=constant$

Therefore, the angular momentum of a particle is conserved when the total force F=0.

answered: Guest

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hope this

answered: Guest

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 $(-\infty,5]$ .