Asimple pendulum consists of a point mass suspended by a weightless, rigid wire in a uniform gravitation field. which of the following statements are true when the system undergoes small oscillations? check all that apply. a. the period is independent of the length of the wire. b. the period is inversely proportional to the length of the wire. c. the period is independent of the suspended mass. d. the period is proportional to the suspended mass. e. the period is proportional to the square root of the length of the wire. f. the period is inversely proportional to the suspended mass.

Option (c) and (e)

Explanation:

In case of a simple pendulum having suspended mass 'M', if the length of the wire is 'L', and the acceleration due to gravity that acts on it is 'g', then the time period, 'T' of the pendulum for one complete oscillation is given by:

From the above equation, we can say that:

Time period of the pendulum does not depend on the suspended mass.Time period is in direct proportion to the square root of the  wire length Time period is depends inversely on the acceleration due to gravity.

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s