A Basketball Player Exerts a Force on a Inflated Basketball
The diameter of the contact between the basketball and the floor
Explanation:
Pressure = force / area
P = F / A
A = F / P
A = 34 N / (9×10⁴ Pa)
A = 0.000378 m²
Area of a circle is:
A = πr²
0.000378 m² = πr²
r = 0.011 m
Diameter is double the radius:
d = 2r
d = 0.022 m
d = 2.2 cm
Final answer:
The diameter of the contact between the basketball and the floor when the player exerts a force of 34N is 1.4 cm, assuming the internal pressure of the ball is evenly distributed.
Explanation:
This Physics problem is essentially asking for a calculation of the diameter resulting from the force applied on the basketball by the player. First, we need to understand that the force applied is equal to pressure times the area (F=P*A). Since we're solving for diameter, we're interested in the area of the contact between the basketball and the floor which is a circle. Therefore, our area will be πr², which results in F=P* πr². Solving for r to get the radius, we have r = sqrt(F/Pπ). We then multiply the radius by 2 to get the diameter.
Here's how it would look with the given numbers:
r = sqrt(34 N / (9 x 10^4 Pa * π)) = 0.00696 m
Diameter = 2r = 0.014 m (or 1.4 cm).
Note that this simplification assumes that the basketball's structure is reasonably uniform and that the pressure inside it is evenly distributed.
What is the relationship between the force applied by the basketball player and the diameter of contact between the basketball and the floor?
The force applied by the basketball player directly affects the diameter of the contact between the basketball and the floor. As the force increases, the diameter of contact also increases. In this scenario, the player exerted a force of 34 N, resulting in a diameter of 2.2 cm.