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3 years ago

In which of the two following cases would you recommend winnowing ?

Physics

1 answer:

fenix001 [56]3 years ago

6 0

${\green{\underline{\textsf{\textbf{Case 2 :When lighter part of a mixture is unwanted and the husk is useful. }}}}}$

${\green{\underline{\textsf{\textbf{Explanation : }}}}}$

As we know that, Winnowing is used when the lighter particles is required to remove. Hence, in Case 2 the lighter part is removed and the remaining is needed.

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3 years ago

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4 years ago

An electric motor spins at 1000 rpm and is slowing down at a rate of 10 t rad/s2 ; where t is measured in seconds. (a) If the mo

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Answer:

a) The tangential component of acceleration at the edge of the motor at $t = 1.5\,s$ is -1.075 meters per square second.

b) The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

Explanation:

The angular aceleration of the electric motor ( $\alpha$ ), measured in radians per square second, as a function of time ( $t$ ), measured in seconds, is determined by the following formula:

$\alpha = -10\cdot t\,\left[\frac{rad}{s^{2}} \right]$ (1)

The function for the angular velocity of the electric motor ( $\omega$ ), measured in radians per second, is found by integration:

$\omega = \omega_{o} - 5\cdot t^{2}\,\left[\frac{rad}{s} \right]$ (2)

Where $\omega_{o}$ is the initial angular velocity, measured in radians per second.

a) The tangential component of aceleration ( $a_{t}$ ), measured in meters per square second, is defined by the following formula:

$a_{t} = R\cdot \alpha$ (3)

Where $R$ is the radius of the electric motor, measured in meters.

If we know that $R = 7.165\times 10^{-2}\,m$ , $\alpha = 10\cdot t$ and $t = 1.5\,s$ , then the tangential component of the acceleration at the edge of the motor is: