The terminal velocity \( v_t \) of a small sphere falling through a viscous medium is given by Stokes' law for low Reynolds numbers: \[ v_t = \frac{2r^2(\rho - \rho_0)g}{9\eta}, \] where: - \( r \) is the radius of the sphere,
- \( \rho \) is the density of the sphere,
- \( \rho_0 \) is the density of the fluid,
- \( g \) is the acceleration due to gravity, and
- \( \eta \) is the dynamic viscosity of the fluid.
From this equation, we can observe that the terminal velocity is directly proportional to the square of the radius of the ball.
Thus, if the radius of the ball increases, the terminal velocity increases with the square of the radius.
Thus, the correct answer is option (C), directly proportional to the square of the radius of the ball.
Let \( f(x) = \frac{x^2 + 40}{7x} \), \( x \neq 0 \), \( x \in [4,5] \). The value of \( c \) in \( [4,5] \) at which \( f'(c) = -\frac{1}{7} \) is equal to:
The general solution of the differential equation \( \frac{dy}{dx} = xy - 2x - 2y + 4 \) is:
The minimum value of the function \( f(x) = x^4 - 4x - 5 \), where \( x \in \mathbb{R} \), is:
The critical points of the function \( f(x) = (x-3)^3(x+2)^2 \) are:
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively: