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:
L = - 1361.591 k Kgm/s
Explanation:
Given
mA = 55.2 Kg
vA = 3.45 m/s
rA = 6.00 m
mB = 62.4 Kg
vB = 4.23 m/s
rB = 3.00 m
mC = 72.1 Kg
vC = 4.75 m/s
rC = - 5.00 m
then we apply the equation
L = (mv x r)
⇒ LA = mA*vA x rA = 55.2 *(3.45 i)x(6 j) = (1142.64 k) Kgm/s
⇒ LB = mB*vB x rB = 62.4 *(4.23 j)x(3 i) = (- 791.856 k) Kgm/s
⇒ LC = mC*vC x rC = 72.1 *(- 4.75 j)x(- 5 i) = (- 1712.375 k) Kgm/s
Finally, the total counterclockwise angular momentum of the three joggers about the origin is
L = LA + LB + LC = (1142.64 - 791.856 -1712.375) k Kgm/s
L = - 1361.591 k Kgm/s
Answer:
141.78 ft
Explanation:
When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.
Calculating the acceleration using one of Newton's equations of motion:

Note: The negative sign denotes deceleration.
When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2
Hence, we can find the minimum stopping distance using:

The minimum stopping distance is 141.78 ft.
I'm not good with math but I think it is 23.4