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
what do you need , i mean your help , let me see if i can help
Answer:
(C) greater than zero but less than 45° above the horizontal
Explanation:
The range of a projectile is given by R = v²sin2θ/g.
For maximum range, sin2θ = 1 ⇒ 2θ = sin⁻¹(1) = 90°
2θ = 90°
θ = 90°/2 = 45°
So the maximum horizontal distance R is in the range 0 < θ < 45°, if θ is the angle above the horizontal.
Linear expansivity, area expansivity and volume or cubic expansivity are
The coefficient of static friction is 0.234.
Answer:
Explanation:
Frictional force is equal to the product of coefficient of friction and normal force acting on any object.
So here the mass of the object is given as 2 kg, so the normal force will be acting under the influence of acceleration due to gravity.
Normal force = mass * acceleration due to gravity
Normal force = 2 * 9.8 = 19.6 N.
And the frictional force is given as 4.6 N, then
Coefficient of static friction = 4.6 N / 19.6 N = 0.234
So the coefficient of static friction is 0.234.