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
Because acceleration is constant, the acceleration of the car at any time is the same as its average acceleration over the duration. So

Now, we have that

so we end up with a distance traveled of


Answer:
The potential difference between the places is 0.3 V.
∴ 1st option i.e. 0.3V is the correct option.
Explanation:
Given
Work done W = 3J
Amount of Charge q = 10C
To determine
We need to determine the potential difference V between the places.
The potential difference between the two points can be determined using the formula
Potential Difference (V) = Work Done (W) / Amount of Charge (q)
or

substituting W = 3 and q = 10 in the formula

V
Therefore, the potential difference between the places is 0.3 V.
∴ 1st option i.e. 0.3V is the correct option.
Distance, since distance represents how far something has travelled, which would be in our case 2.5m.
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The moment of the resultant of these two forces with respect to O 376 lb-ft CCW which is <span>about moment center point O.</span>