The answer is B, because it will lose potential energy.
Answer:
Option B:
A child sitting on a swing.
Explanation:
When we hear the word oscillator, a good example is the pendulum bob of a grandfather clock. We can picture the motion to get a perfect understanding of its path of motion and relate it to other systems of motion in our everyday life.
An oscillator is a system that moves in such a way that it reverses its direction after a period of time. It can be seen as a "to-and-fro" motion.
From the options, a child sitting on a swing is the perfect example of an oscillating system because the child will be moving forwards and backwards, alternately reversing the direction of motion with time.
Answer:
1. The length is 8.35m
2. The period on the moon is 14.05 secs
Explanation:
1. Data obtained from the question. This includes the following:
Period (T) = 5.8 secs
Acceleration due to gravity (g) = 9.8 m/s2
Length (L) =...?
The length can be obtained by using the formula given below:
T = 2π√(L/g)
5.8 = 2π√(L/9.8)
Take the square of both side
(5.8)^2 = 4π^2 x L/ 9.8
Cross multiply
4π^2 x L = (5.8)^2 x 9.8
Divide both side by 4π^2
L = (5.8)^2 x 9.8 / 4π^2
L= 8.35 m
2. Data obtained from the question. This includes the following:
Acceleration due to gravity (g) = 1.67 m/s2
Length (L) = 8.35m (the length remains the same)
Period (T) =?
The period can be obtained as follow:
T = 2π√(L/g)
T = 2π√(8.35/1.67)
T = 14.05 secs
Therefore, the period on the moon is 14.05 secs
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
Acceleration due to gravity
