Physics, 27.08.2020 15:01 missmychiefrkt0

A particle makes 240 revolution per minutes on a circular radius of 2m find: (I) it's period (ii) it's angular velocity (III) it's linear velocity and (iv) it's acceleration

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(i) The period is 0.25 seconds, (ii) The angular velocity is approximately 25.133 radians per second, (iii) The linear velocity of the particle is approximately 50.266 meters per second, (iv) The acceleration experimented by the particle is approximately 1263.336 meters per square second.

Explanation:

(i) Let suppose that particle experiments an uniform circular motion. The period () of the particle, measured in seconds, is the time needed to complete a revolution, which is determined by this:

$T = \frac{2\pi}{\omega}$

Where $\omega$ is the angular speed, measured in radians per second.

Let is convert angular speed from revolutions per minute into radians per second:

$\omega = \left(240\,\frac{rev}{min} \right)\times \left(\frac{1}{60}\,\frac{min}{s} \right)\times \left(2\pi\,\frac{rad}{rev} \right)$

$\omega \approx 25.133\,\frac{rad}{s}$

The period of the particle is:

$T = \frac{2\pi}{25.133\,\frac{rad}{s} }$

$T \approx 0.25\,s$

The period is 0.25 seconds.

(ii) The angular velocity is approximately 25.133 radians per second.

(iii) The linear velocity of the particle (), measured in meters per second, is found by this expression:

$v = R\cdot \omega$

Where:

- Radius, measured in meters.

$\omega$ - Angular velocity, measured in radians per second.

If $R = 2\,m$ and $\omega \approx 25.133\,\frac{rad}{s}$ , the linear velocity of the particle is:

$v = (2\,m)\cdot \left(25.133\,\frac{rad}{s} \right)$

$v \approx 50.266\,\frac{m}{s}$

The linear velocity of the particle is approximately 50.266 meters per second.

(iv) As particle rotates at constant speed, resultant acceleration () is entirely radial and modelled by this expression:

$a = \omega^{2}\cdot R$

$a = \left(25.133\,\frac{rad}{s} \right)^{2}\cdot (2\,m)$

$a \approx 1263.336\,\frac{m}{s^{2}}$

The acceleration experimented by the particle is approximately 1263.336 meters per square second.

Answer from: Quest

reserve is the answer

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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A particle makes 240 revolution per minutes on a circular radius of 2m find: (I) it's period (ii) it...