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


Answer:
The net displacement of the car is 3 km West
Explanation:
Please see the attached drawing to understand the car's trajectory: First in the East direction for 4 km (indicated by the green arrow that starts at the origin (zero), and stops at position 4 on the right (East).
Then from that position, it moves back towards the West going over its initial path, it goes through the origin and continues for 3 more km completing a moving to the West a total of 7 km. This is indicated in the drawing with an orange trace that end in position 3 to the left (West) of zero.
So, its NET displacement considered from the point of departure (origin at zero) to the final point where the trip ended, is 3 km to the west.
It is called surface tension it is the elastic personality of some liquids as they pull together to take up as little surface area as possible. the water molecules would rather stay together than be pulled apart<span />
Answer:
The tangential speed of the tack is 6.988 meters per second.
Explanation:
The tangential speed experimented by the tack (
), measured in meters per second, is equal to the product of the angular speed of the wheel (
), measured in radians per second, and the distance of the tack respect to the rotation axis (
), measured in meters, length that coincides with the radius of the tire. First, we convert the angular speed of the wheel from revolutions per second to radians per second:


Then, the tangential speed of the tack is: (
,
)


The tangential speed of the tack is 6.988 meters per second.
Answer:
The bullet that is fired will spend longer in the air, hitting the ground after the dropped bullet.
Explanation:
Using the equation: x
= x
0 +
v
t
If we neglect the effects of air resistance, the horizontal motion is a constant velocity.
The horizontal displacement = (velocity X cosθ)
So, the fired bullet has to travel horizontally before falling which takes a longer time compared to a bullet dropped where it is, height = 1/2 gt^2
gravity, g = 9.8 m/s2.