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:
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Answer:
The value is 
Explanation:
From the question we are told that
The mass of matter converted to energy on first test is 
The mass of matter converted to energy on second test 
Generally the amount of energy that was released by the explosion is mathematically represented as
=> ![E = 1.5 *10^{-3} * [ 3.0 *10^{8}]^2](https://tex.z-dn.net/?f=E%20%3D%20%201.5%20%2A10%5E%7B-3%7D%20%20%2A%20%5B%203.0%20%2A10%5E%7B8%7D%5D%5E2)
=>
