7. An object falling in air experiences a drag force directly related to the square of the speed such that F = Cv2 (where C is a constant of proportionality). Assuming that the buoyant force due to air is negligible, the terminal velocity of this falling body is best described by the equation:
(The terminal velocity is the speed at which the object is in equilibrium).
a.) mg/C
b.) [�ã (mg)] / C
c.) mg / �ã (C)
d.) �ã (mg/C)
e.) C/mg
e.) C/mg
Why did the object become a falling body in the first place? Did it miss the bus or did it trip on its shoelaces?
The terminal velocity of the falling body occurs when the drag force is equal to the gravitational force acting on the object.
The gravitational force acting on the object is given by:
F_gravity = m * g
The drag force is given by:
F_drag = C * v^2
At terminal velocity, F_gravity = F_drag.
Therefore,
m * g = C * v^2
To find v (the terminal velocity), we can solve for it:
v^2 = (m * g) / C
Taking the square root of both sides:
v = sqrt((m * g) / C)
Therefore, the terminal velocity of the falling body is given by option d.) √(mg/C).
To find the terminal velocity of a falling object, we need to set the drag force equal to the weight of the object. The drag force can be expressed as F = Cv^2, where C is a constant and v is the velocity (speed) of the falling object.
The weight of the object is given by the formula W = mg, where m is the mass of the object and g is the acceleration due to gravity.
Setting the drag force equal to the weight, we have:
Cv^2 = mg
To solve for v, we can isolate it by dividing both sides of the equation by C:
v^2 = mg/C
Taking the square root of both sides, we get:
v = √(mg/C)
So the terminal velocity, v, is equal to the square root of (mg/C). Therefore, the correct option that describes the terminal velocity of the falling body is:
d.) √(mg/C)