a. 0.5 T
- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position
- The period T is the time the system takes to complete one oscillation
During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.
So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

and solving for t we find

b. 1.25T
Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that
- the mass takes a time of 1 T to cover a distance of 4A
we can set the following proportion:

And by solving for t, we find

Answer:
Explanation:
Relative velocity is defined as the velocity of an object B in the rest frame of another object A.
<span>F = m*g = 88.6kg * 9.81 m/s^2 = 869 N (3 s.f.)
The equation above is Newton's 2nd Law of Motion where g is the acceleration due to gravity near the Earth's surface.</span><span>
</span>
The net force acting on the center O is 4.495 X 10⁹ N
<u>Explanation:</u>
Given:
Charge-
q₁ = 4C
q₂ = 3C
q₃ = -5C
q₄ = 2C
q₅ = 1C
Side of the square = 4cm
Net force acting on the center, F = ?
Length of the side of the square is 4cm
The length of the diagonal will be 4√2
The length of half diagonal = 4√2 / 2 = 2√2
Force acting on the center of the square is:


Here, k is coulomb's constant and the value is
k = 8.99 x 10⁹ N m2 / C2
On substituting the value, we get:

Therefore, the net force acting on the center O is 4.495 X 10⁹ N