Respuesta :
Answer:
3.71 m/s
Explanation:
Given data :
radius of the spherical ball = 0.400 m
Density of the ball material = 10.65 Kg/m^3
Density of water = 1000 Kg/m^3
Drag coefficient C_d= 0.520
Constant upward force F_up= 2630
Concept: The moment before it starts to rise Upward force must be equal to downward drag force
⇒ F_up = F_D
F_D =C_d× 0.5 ×ρ_w×V^2×A
where V= terminal velocity and A= area πR^2
equating the forces we get
2630= 0.52× 0.5×1000×V^2×π(0.4)^2
V= 3.171 m/s
Therefore, the terminal velocity of the ball = 3.71 m/s
The ball while moving through the water in the pool, experiences a drag force. Then, the magnitude of terminal speed during the rise of the ball is 3.171 m/s.
What is terminal speed?
When an object is made to fall through a liquid medium, then the maximum velocity achieved at that instant is known as terminal velocity or terminal speed.
Given data-
The radius of the spherical beach ball is, r = 0.4000 m.
The density of the ball is, [tex]\rho'=10.65 \;\rm kg/m^{3}[/tex].
The density of water in the pool is, [tex]\rho =1000.0 \;\rm kg/m^{3}[/tex].
The drag coefficient of the ball in water is, [tex]C_{d}=0.520[/tex].
The magnitude of the constant upward force is, F = 2630 N.
As per the concept of Buoyancy, "The moment before the ball starts to rise upward force must be equal to downward drag force".
F = Fd
Here,
F is the upward force.
Fd is the downward force.
[tex]F=\dfrac{1}{2} \times C_{d} \times \rho \times v^{2} \times A[/tex]
Here,
A is the area of the ball.
Solving as,
[tex]2630=\dfrac{1}{2} \times 0.520 \times 1000 \times v^{2} \times (\pi r^{2})\\\\2630=\dfrac{1}{2} \times 0.520 \times 1000 \times v^{2} \times (\pi \times 0.4000^{2})\\\\v=3.171 \;\rm m/s[/tex]
Thus, we can conclude that the magnitude of terminal speed during the rise of the ball is 3.171 m/s.
Learn more about the terminal speed here:
https://brainly.com/question/26505185
