Given v_in = 20 m/s and a = 3 m/s2, assuming that the body
moves at constant acceleration, the motion is modeled by the equation:
s(t) = (v_in)t + (1/2)a(t^2)
where s(t) is the distance traveled
substituting the given,
s(t) = 20t + (3/2)(t^2)
at t = 3
s(t) = 20(3) + (3/2)(3)^2
= 73.5 m
Answer:
Explanation:
The force experimented by a charge <em>q </em>in a uniform electric field <em>E</em><em> </em>is <em>F=qE</em>.
Newton's 2nd Law tells us that the relation between acceleration <em>a</em> a mass <em>m </em>experiments when a force <em>F </em>is applied to it is <em>F=ma</em>.
Combining these equations we have <em>am=qE</em>, and since we want the acceleration of the speck of dust, we substitute our values:
You could hold any object (like an apple) for your class to see. (Its potential energy is greatest at this point). At the point when you are holding the object the potential energy will be equal to the object's mass multiplied by the object's acceleration due to gravity(9. 8 m/s²) multiplied by the height of the object(however high you choose to hold it). Release the object while it is falling, the object's motion will be evidence of the kinetic energy that the object is experiencing. As the object's kinetic energy increases, its potential energy will decrease. This can be explained by the law of conservation of energy. This law states that energy cannot be created or destroyed it can only change forms. Finally, explain to your class that mechanical energy is the sum of kinetic and potential energy.
I hope this helped. I recommend you present with an informative powerpoint in the background of your presentation while you present this if you want to do well because it will better show your teacher how much you know rather than if you were to just speak to the class during your presentation.
The starting angle θθ of a pendulum does not affect its period for θ<<1θ<<1. At higher angles, however, the period TT increases with increasing θθ.
The relation between TT and θθ can be derived by solving the equation of motion of the simple pendulum (from F=ma)
−gsinθ=lθ¨−gainθ=lθ¨
For small angles, θ≪1,θ≪1, and hence sinθ≈θsinθ≈θ. Hence,
θ¨=−glθθ¨=−glθ
This second-order differential equation can be solved to get θ=θ0cos(ωt),ω=gl−−√θ=θ0cos(ωt),ω=gl. The period is thus T=2πω=2πlg−−√T=2πω=2πlg, which is independent of the starting angle θ0θ0.
For large angles, however, the above derivation is invalid. Without going into the derivation, the general expression of the period is T=2πlg−−√(1+θ2016+...)T=2πlg(1+θ0216+...). At large angles, the θ2016θ0216 term starts to grow big and cause
