A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should be unbinding the strings to achieve a speed of 4 m/s, is ______ cm. (Take g = 10 m/s2).
The correct answer is 120
\(α = \frac{(mg)(r)}{\frac{3}{2}mr^2} = \frac{2g}{3r}\)
\(⇒ a = \frac{2g}{3}\)
\(⇒ v^2 = 2as\)
\(16 = \frac{40}{3}×s\)
⇒ s = 0.3×4
= 120cm
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
Rotational motion can be defined as the motion of an object around a circular path, in a fixed orbit.
The wheel or rotor of a motor, which appears in rotation motion problems, is a common example of the rotational motion of a rigid body.
Other examples: