Answer: YES !!
Explanation: THAT WOULD BE AMAZING!!!
Answer:
A) 35 ft
B) 5 ft
C) Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick
Explanation:
A) Total distance covered by the dog = 20 + 15
= 35 ft
B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;
total displacement of the dog = 20 - 15
= 5 ft
C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick
But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.
Thus,
Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick
As we know that KE and PE is same at a given position
so we will have as a function of position given as
also the PE is given as function of position as
now it is given that
KE = PE
now we will have
so the position is 0.707 times of amplitude when KE and PE will be same
Part b)
KE of SHO at x = A/3
we can use the formula
now to find the fraction of kinetic energy
now since total energy is sum of KE and PE
so fraction of PE at the same position will be
If we are being specific, the inner core has the highest density, but if not then the core in general
S = ut + 0.5at^2
<span>10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414} </span>
<span>t^2 ~ 2.04 </span>
<span>t ~ 1.43 seconds</span>
