There are competitions in which pilots fly small planes low over the ground and drop weights, trying to hit a target. A pilot fl
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Answer:
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
The mechanical energy for the first and the second ball is
Mass of the first ball = 20 kg
The initial speed at which a cannonball is fired from a cannon =1000 m/s
The angle made by the cannonball while being fired from the cannon = 37°
The maximum height reached by the first ball is,
The maximum height of the first cannonball is 17478.69 m.
The initial speed at which a cannonball is fired from a cannon =1000 m/s
The angle made by the cannonball while being fired from the cannon = 90 °
For the first ball, total mechanical energy= Potential energy at maximum height + kinetic energy at the maximum height
So, the total mechanical energy is,
= mgh \: + \frac{1}{2}mv {}^{2} _{x}[/tex]
Therefore, the total mechanical energy for the first and the
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