Physics, 16.01.2020 07:31, Teenabrewer28

Arotating wheel has a constant angular acceleration. it has an angular velocity of 5.0 rad/s at time t = 0 s, and 3.0 s later has an angular velocity of 9.0 rad/s. what is the angular displacement of the wheel during the 3.0-s interval?

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answered: juniorcehand04

Angular displacement during 3 seconds is 21 rad

Explanation:

We have equation of motion v = u + at

Initial velocity, u = 5 rad/s

Final velocity, v = 9 rad/s

Time, t = 3 s

Substituting

v = u + at

9 = 5 + a x 3

a = 1.33 rad/s²

Angular acceleration is 1.33 rad/s²

Now we have equation of motion v² = u² + 2as

Initial velocity, u = 5 rad/s

Final velocity, v = 9 rad/s

Angular acceleration = 1.33 rad/s²

Substituting

v² = u² + 2as

9² = 5² + 2 x 1.33 x s

s = 21 radians

Angular displacement during 3 seconds is 21 rad

answered: Guest

Answer

the order for strong to weak forces such as

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

explanation

the fundamental forces are those forces that act between to objects or element to make interactions. there are four types of fundamental forces such as strong and weak nuclear forces (due to protons and electrons), electromagnetic force (due to the motion of electrons) and gravitational force (due to orbital movement of the planet). if these forces are arranged on the basis of their maximum strength then the order is

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

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 $(-\infty,5]$ .

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Arotating wheel has a constant angular acceleration. it has an angular velocity of 5.0 rad/s at time...