most of the earth is covered with oceans
Answer:
Incomplete question
The complete question is
A Ferris wheel is a vertical, circular amusement ride with radius 6.0 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 9.6 s. Consider a rider whose mass is 96 kg.
At the bottom of the ride, what is the rate of change of the rider's momentum?
Explanation:
Radius of wheel is 6m
Rider mass=96kg
He completes one revolution in 9.6s
Let get angular velocity (w)
1 Revolution =2πrad
θ=2πrad
w= θ/t
w=2π/9.6
w=0.654rad/s
Linear speed is give as
v=wr
v=0.654×6
v=3.93m/s
Centripetal acceleration a
a=rw²
a=6×0.654²
a=2.57m/s²
Acceleration due to gravity g=9.81m/s²
According to Newton's second law of motion net force acting on the rider at the bottom of the ride is given by: the two force acting at the bottom is the normal and the weight of the rider
ΣF = ma
N-W=ma
N-mg=ma
N=ma+mg
N=m(a+g)
N=96(2.57+9.81)
N=1188.48 N
Therefore the rate of change of momentum at the bottom of the ride is 1188.48 N.
Answer:
so the speed will increase by 1.44 times then the initial speed if the distance is increased to double
Explanation:
As we know that the air friction or resistance due to air is neglected then we can use the equation of kinematics here
since we released it from rest so we have
so here we have
now if the distance is double then we have
now from above two equations we can say that
so the speed will increase by 1.44 times then the initial speed if the distance is increased to double
Answer:
7.5 m/s²
Explanation:
Given:
v₀ = 0 m/s
v = 30 m/s
t = 4 s
Find: a
v = at + v₀
(30 m/s) = a (4 s) + (0 m/s)
a = 7.5 m/s²
