An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.220 rev/s . The magnitude of the angular acceleration is 0.920 rev/s2 . Both the the angular velocity and angular acceleration are directed counterclockwise. The electric ceiling fan blades form a circle of diameter 0.800 m . 1) Compute the fan's angular velocity magnitude after time 0.208 s has passed.
2)Through how many revolutions has the blade turned in the time interval 0.208 s from Question 1?
3)What is the tangential speed vt of a point on the tip of the blade at time t = 0.208 s ?
4)Calculate the magnitude at of the tangential acceleration of a point on the tip of the blade at time t = 0.208 s .
5)Calculate the magnitude ar of the radial (or centripetal) acceleration of the point at the end of the fan blade.

1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s

2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²

Explanation:

The given parameters are;

The initial velocity of the fan, n = 0.220 rev/s

The magnitude of the angular acceleration = 0.920 rev/s²

The direction of the angular acceleration and the angular velocity = Clockwise

The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m

1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s

The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²

The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;

ω = ω₀ + α·t

From which we have;

ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s

The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s

2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;

θ = ω₀·t + 1/2·α·t²

θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians

2·π radians = 1 revolution

∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution

The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;

The tangential speed, = ω × r = ω × D/2

At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;

= ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413

The tangential speed, = 1.0338604413 m/s

The tangential speed ≈ 1.034 m/s

4)  The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;

a = α × r = α × D/2

a = 5.78053048261 × 0.800/2 = 2.31221219304

The tangential acceleration, a ≈ 2.312 m/s²

answer: d. a cme increases the amount of energy in the solar wind

explanation:

coronal mass ejections, also called cmes, are large clouds of plasma and magnetic field that erupt from the sun. only when the cloud is aimed at earth will the cme hit earth and therefore cause impacts. high-speed solar wind streams come from areas on the sun known as coronal holes.