The first law is about force or push and pull
Answer:
A = 2.36m/s
B = 3.71m/s²
C = 29.61m/s2
Explanation:
First, we convert the diameter of the ride from ft to m
10ft = 3m
Speed of the rider is the
v = circumference of the circle divided by time of rotation
v = [2π(D/2)]/T
v = [2π(3/2)]/4
v = 3π/4
v = 2.36m/s
Radial acceleration can also be found as a = v²/r
Where v = speed of the rider
r = radius of the ride
a = 2.36²/1.5
a = 3.71m/s²
If the time of revolution is halved, then radial acceleration is
A = 4π²R/T²
A = (4 * π² * 3)/2²
A = 118.44/4
A = 29.61m/s²
Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
It’s a loose spherical structure which is located around the bulge and some of the disk
A surface wave is a wave in which particles of the medium undergo a circular motion.