First we need to know the equation:
F= Mass times acceleration
F = 2.0 kg times 5.0 m/s^2
multiply them to get the net force!
F= 10 N
N is newton
Hope this heps
Answer:
<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2>

</h2>
Explanation:
Part a)
As we know that drag force is given as






so we have


So acceleration of the ball is



Part B)
As per kinematics we know that



Answer:
Exercise 1;
The centripetal acceleration is approximately 94.52 m/s²
Explanation:
1) The given parameters are;
The diameter of the circle = 8 cm = 0.08 m
The radius of the circle = Diameter/2 = 0.08/2 = 0.04 m
The speed of motion = 7 km/h = 1.944444 m/s
The centripetal acceleration = v²/r = 1.944444²/0.04 ≈ 94.52 m/s²
The centripetal acceleration ≈ 94.52 m/s²
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