<span>Radius, the distance from the centre = 0.390
Electric field is equal to half of the magnitude. E2 = E / 2
Given
E1 = E2
E1 = k x Q / r^2
E2 = (k x Q / r2^2) / 2
Equating the both we get 2 x r^2 = r2^2
r2 = square root of (2 x r1^2) = square root of (2) x r = 1.414 x 0.390
r2 = 1.414 x 0.390 = 0.551 m</span>
Answer:
H(max) = (v²/2g)
Explanation:
The maximum height the ball will climb will be when there is no friction at all on the surface of the hill.
Normally, the conservation of kinetic energy (specifically, the work-energy theorem) states that, the change in kinetic energy of a body between two points is equal to the work done in moving the body between the two points.
With no frictional force to do work, all of the initial kinetic emergy is used to climb to the maximum height.
ΔK.E = W
ΔK.E = (final kinetic energy) - (initial kinetic energy)
Final kinetic energy = 0 J, (since the body comes to rest at the height reached)
Initial kinetic energy = (1/2)(m)(v²)
Workdone in moving the body up to the height is done by gravity
W = - mgH
ΔK.E = W
0 - (1/2)(m)(v²) = - mgH
mgH = mv²/2
gH = v²/2
H = v²/2g.
We have that the Throw has changed in distance the ball has traveled and the Force applied in trowing the ball and Possibly the time of travel
From the Question we are told that
You toss a ball to someone one meter away
You toss it to someone four meters away
Generally
When you toss toss a ball to someone one meter away and You toss it to someone four meters away the are things that have changed
Therefore
The Throw has changed in distance the ball has traveled and the Force applied in trowing the ball and Possibly the time of travel
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