Answer:
Speed of the aircraft = 36.64 m/s
Explanation:
Consider the vertical motion of the projectile,
We have equation of motion s = ut+0.5at²
Let the velocity of plane be v.
Vertical velocity = vsin55 = Initial velocity of projectile in vertical direction = u
acceleration, a = 9.81 m/s²
displacement , s = 554 m
time, t = 8 s
Substituting,
554 = vsin55 x 8 +0.5 x 9.81 x 8²
v = 36.64 m/s
Speed of the aircraft = 36.64 m/s
Answer:
4.36 rad/s
Explanation:
Radius of platform r = 2.97 m
rotational inertia I = 358 kg·m^2
Initial angular speed w = 1.96 rad/s
Mass of student m = 69.5 kg
Rotational inertia of student at the rim = mr^2 = 69.5 x 2.97^2 = 613.05 kg.m^2
Therefore initial rotational momentum of system = w( Ip + Is)
= 1.96 x (358 + 613.05)
= 1903.258 kg.rad.m^2/s
When she walks to a radius of 1.06 m
I = mr^2 = 69.5 x 1.06^2 = 78.09 kg·m^2
Rotational momentuem of system = w(358 + 78.09) = 436.09w
Due to conservation of momentum, we equate both momenta
436.09w = 1903.258
w = 4.36 rad/s
Answer: τ = 0
Explanation:
At constant angular velocity there is no angular acceleration therefore no torque.
τ = Iα
