What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 45 rpm (revolutions per minute) if the wheel's diameter is 40 cm ?

Respuesta :

The magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 45 rpm (revolutions per minute) if the wheel's diameter is 40 cm is [tex]4.418m/s^2[/tex].

Acceleration and change in velocity has both magnitude and acceleration because it is a vector quantity. Units are used to describe magnitude. The factors that affect acceleration include time, velocity, force, and many others. Both translation and rotation are possible.

The vector's magnitude determines its length. The vector's direction determines its direction. As a result, the acceleration's magnitude matches the acceleration vector's magnitude, and its direction matches the acceleration vector's direction.

Acceleration of speck of clay

[tex]a_R=V^2/R[/tex]

[tex]a_R=(\omega R^2)/R[/tex]

[tex]a_R=\omega R[/tex]

Given, the frequency of Potters wheel

f=45rpm

f=45(1min/60sec)

f=45/60 revolution/sec

ω=2πf

ω=2(3.14)(45/60)

ω=4.7 rad/sec

[tex]a_R=(4.7 rad/sec)^2(0.40/2 m)[/tex]

[tex]a_R=4.418m/s^2[/tex]

To learn more about magnitude of the acceleration refer the link:

https://brainly.com/question/15015453

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