What is the peacocks displacement from 2 to 3 seconds

Physics

1 answer:

mash [69]3 years ago

7 0

1.5 second difference

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2. a) A disc rotates about its axis at speed 25 revolutions per minute and takes 15 s to stop. Calculate the

ASHA 777 [7]

The statement shows a case of rotational motion, in which the disc <em>decelerates</em> at <em>constant</em> rate.

i) The angular acceleration of the disc ( $\alpha$ ), in revolutions per square second, is found by the following kinematic formula:

$\alpha = \frac{\omega_{f}-\omega_{o}}{t}$ (1)

Where:

$\omega_{o}$ - Initial angular speed, in revolutions per second.
$\omega_{f}$ - Final angular speed, in revolutions per second.
$t$ - Time, in seconds.

If we know that $\omega_{o} = \frac{5}{12}\,\frac{rev}{s}$ , $\omega_{f} = 0\,\frac{rev}{s}$ y $t = 15\,s$ , then the angular acceleration of the disc is:

$\alpha = \frac{0\,\frac{rev}{s}-\frac{5}{12}\,\frac{rev}{s}}{15\,s}$

$\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}$

The angular acceleration of the disc is $\frac{1}{36}$ radians per square second.

ii) The number of rotations that the disk makes before it stops ( $\Delta \theta$ ), in revolutions, is determined by the following formula:

$\Delta \theta = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \alpha}$ (2)

If we know that $\omega_{o} = \frac{5}{12}\,\frac{rev}{s}$ , $\omega_{f} = 0\,\frac{rev}{s}$ y $\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}$ , then the number of rotations done by the disc is:

$\Delta \theta = 3.125\,rev$