At the lowest point on the Ferris wheel, there are two forces acting on the child: their weight of 430 N, and an upward centripetal/normal force with magnitude n; then the net force on the child is
∑ F = ma
n - 430 N = (430 N)/g • a
where m is the child's mass and a is their centripetal acceleration. The child has a linear speed of 3.5 m/s at any point along the path of the wheel whose radius is 17 m, so the centripetal acceleration is
a = (3.5 m/s)² / (17 m) ≈ 0.72 m/s²
and so
n = 430 N + (430 N)/g (0.72 m/s²) ≈ 460 N
Answer:
(a) Yes, it is possible by raising the object to a greater height without acceleration.
Explanation:
The work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in kinetic energy requires a change in velocity.
If kinetic energy will not change, then velocity will not change, this means that there will be constant velocity and an object with a constant velocity is not accelerating.
If the object is not accelerating (without acceleration) and it remains at the same height (change in height = 0, and mgh = 0).
Thus, for work to be done on the object, without changing the kinetic energy of the object, the object must be raised to a greater height without acceleration.
Correct option is " (a) Yes, it is possible by raising the object to a greater height without acceleration".
Explanation:
may I ask what it is on because I might be able to help out
Answer:
mean = 10.68 m/s
standard deviation 0.3059
[/tex]\sigma_m = 0.14[/tex]
Explanation:
1) 
mean = 10.68 m/s
2 ) standard deviation is given as
N = 5
SOLVING ABOVE RELATION TO GET STANDARD DEVIATION VALUE
\sigma = 0.3059
3) ERROR ON STANDARD DEVIATION
