At the instant its angular displacement is 0.32 rad, the angular acceleration of a physical pendulum is -630 rad/s2. What is its
2 answers:
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Answer:
The true course: north of east
The ground speed of the plane: 96.68 m/s
Explanation:
Given:
= velocity of wind =
= velocity of plane in still air =
Assume:
= resultant velocity of the plane
= direction of the plane with the east
Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.
Let us find the direction of this resultant velocity with respect to east direction:
This means the the true course of the plane is in the direction of north of east.
The ground speed will be the magnitude of the resultant velocity of the plane.
Hence, the ground speed of the plane is 96.68 km/h.
Answer:
4.15 m/s
Explanation:
As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:
ΔE = ΔK + ΔU =0
If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:
h₀ = 1.09 m -0.21 m = 0.88 m
⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J
As Uf = 0, ΔU = Uf -U₀ = -352 J
As the swing starts from rest, K₀=0, so we can say:
ΔK = Kf = (1)
As ΔK = -ΔU ⇒ ΔK = 352 J (2)
From (1) and (2) we can solve for vf, as follows:
So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.