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
Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?
B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>
<span>Solution:
A) A charge q </span>under an electric field of intensity E will experience a force F equal to:

In our problem we have
and
, so we can find the magnitude of the electric field:

The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.
B) The proton charge is equal to

Therefore, the magnitude of the force acting on the proton will be

And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.
Answer:
10.23m/s^2
Explanation:
GIven data
mass of elevator = 2125 kg
Force= 21,750 N
Required
The maximum acceleration upward
F= ma
a= F/m
a=21,750/2125
a= 10.23m/s^2
Hence the acceleration is 10.23m/s^2
If a bus travels 30 km in 1/2 hr, then in one hr, he can travel twice the distance.
30*2=60 km
Final answer: 60 km per hr