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:
can u try to calculate it and you will get the answer
You would have to give it more mechanical energy.
Like, strap a bunch of powerful rockets to one side of the moon, with all of them pointing in the direction that the moon is already moving in its orbit. Then blast away.
NOTE: There aren't enough rockets or rocket fuel on Earth to make a difference, even if you used ALL of them. The mass of the moon is about
<em>73,476,730,900,000,000,000,000 kilograms</em>
(rounded to the nearest hundred trillion kilograms.)
That's a lot.
Answer:
Explanation:
Electric field between plates
V / d
= 170 / ( 2 x 10⁻² )
= 8500 N/C
Force on electron in this field
= 8500 x 1.6 x 10⁻¹⁹
= 13600 x 10⁻¹⁹ N
Acceleration
= 13600 x 10⁻¹⁹ / 9.1 x 10⁻³¹
a = 1494.5 x 10¹² m /s²
s = .1 x 10⁻² m
v² = u² + 2as
= (2.9x 10⁵)²+ 2 x 1494.5 x 10¹² x .1 x 10⁻²
= 8.41 x 10¹⁰ + 299 x 10¹⁰
= (8.41 + 299 ) x 10¹⁰
v = 17.53 x 10⁵ m /s
