Answer:
R = 1.8 m
Explanation:
This is a simple harmonic movement exercise, at the bottom of the swing the acceleration is vertical upwards and the speed is tangential to the trajectory, that is horizontal; the expression for the centralized acceleration is
= v² / R
R = v² /a_{c}
where the radius is equal to the length of the swing
let's calculate
R = 8.1 / 4.5
R = 1.8 m
Answer:
Car A has twice the energy of car B.
Explanation:
The kinetic energy of an object is given by :
Where
m is the mass
v is the speed of the object
Car A has twice the mass of Car B. Both travel at the same speed i.e.
So,
So, the kinetic energy of car A is twice of the kinetic energy of car B.
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