At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball's acceleration is 810 m/s2 and the vertical or y component of its acceleration is 910 m/s2. The ball's mass is 0.32 kg. What is the magnitude of the net force acting on the soccer ball at this instant?

X = 810 m/s^2

Y = 910 m/s^2
m = 0.32 kg
F = ?

a^2 = X^2 + Y^2 = 810^2 + 910^2 =
1,484,200
a = 1218 m/s2.

F = m*a = 0.32 * 1218 = 390 N

To find the magnitude of the net force acting on the soccer ball, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.

In this case, we have the mass of the ball (m = 0.32 kg) and the components of its acceleration (ax = 810 m/s^2, ay = 910 m/s^2).

The net force can be divided into its horizontal (Fx) and vertical (Fy) components, where Fx is equal to mass times the horizontal acceleration (Fx = m * ax) and Fy is equal to mass times the vertical acceleration (Fy = m * ay).

So we have:
Fx = 0.32 kg * 810 m/s^2 = 259.2 N
Fy = 0.32 kg * 910 m/s^2 = 291.2 N

To find the magnitude of the net force (F), we can use the Pythagorean theorem, which states that the square of the hypotenuse (F) of a right triangle is equal to the sum of the squares of the other two sides (Fx and Fy):

F^2 = Fx^2 + Fy^2

Substituting the values we calculated earlier:
F^2 = (259.2 N)^2 + (291.2 N)^2

Calculating:
F^2 = 67052.64 N^2 + 84639.04 N^2
F^2 = 151691.68 N^2

Finding the square root of both sides:
F = √151691.68 N^2 ≈ 389.62 N

Therefore, the magnitude of the net force acting on the soccer ball at this instant is approximately 389.62 Newtons.