a wheel 0.35m in diameter rotates at 2200rpm. calculate its angular velocity in rad/s and its linear speed and acceleration of a point on the edge of the wheel. need help asap

( Angular Velocity = ( About ) 230 rad / s,

( Linear Speed = ( About ) 40.25 m / s,

( Acceleration = ( About ) 9290 m /

Explanation:

Here we want the angular velocity in radians per second, the linear velocity and acceleration.

The diameter = .35 meters, and thus we can conclude that the radius be half of that, or .175 meters. For part ( a ), or the calculation of the angular velocity, it is given that the diameter rotates at 2200 revolutions per minute - but we need to convert this into radians per second.

We can say that there are 2π radians for every minute, and for every minute there are 60 seconds. Therefore -

( a ) w = 2,200 rpm( 2π rads / rev )( 1 min / 60 sec )...

Hence, w = ( About ) 230 rad / s

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For this second part we can calculate the the linear velocity by multiplying the angular velocity ( omega ) by the radius r -

( b ) v = w( r ) - Substitute,

v = ( 230 rad / sec )( .175 m )...

v = ( About ) 40.25 m / s

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And for this last bit here, to find the acceleration we can simply take the angular velocity ( omega ) squared, by the radius r -

( c ) = w^2( r ),

= ( ( 230 rad / sec )^2 )( .175 m )...

= ( About ) 9290 m /

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

-reverse card

c) up is the direction the magnetic force is acting on the charge.