When precipitation hits the ground and hydrosphere
The magnitude of the E-field decreases as the square of the distance from the charge, just like gravity.
Location ' x ' is √(2² + 3²) = √13 m from the charge.
Location ' y ' is √ [ (-3)² + (-2)² ] = √13 m from the charge.
The magnitude of the E-field is the same at both locations.
The direction is also the same at both locations ... it points toward the origin.
Based on Newton's second law of motion, the net force applied to an object is equal to the product of the mass of the object and the acceleration it experiences. That is,
F = ma
If we are to assume that the net force is constant and that the mass is increased, the acceleration should therefore decrease in order to make constant the value at the right-hand side of the equation.
The AMA is calculated as:
AMA = force obtained / force applied
AMA = 400 / 40
AMA = 10
Answer: Option B. R = (1/2)gt^2
Explanation:
S = R (horizontal distance)
V^2 = 2gS
V^2 = 2gR
R = V^2 / 2g
But V = gt
R = (gt)^2 / 2g
R = (g^2 x t^2) / 2g
R = gt^2 / 2
But t^2 = 2h/g
R = ( g x 2h/g) / 2
R = h
But h = (1/2)gt^2
R = h = (1/2)gt^2
