The answer to this statement is true!
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:0 J
Explanation:
Given
For first step
change in internal Energy of the system is 
Work done on the system 
For second step
change in internal Energy of the system is 
Work done on the system 
Work done on the system is considered as Positive and vice-versa.
and from first law of thermodynamics

for first step


overall heat added
For overall Process Heat added is 0 J
I believe all of these would be known as specific phobias.
Answer:
a) x(t) = 10t + (2/3)*t^3
b) x*(0.1875) = 10.18 m
Explanation:
Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.
Given:
- v(t) = 10 + 2*t^2 (radar gun)
- x*(t) = 10 + 5t^2 + 3t^3 (our coordinate)
Find:
-The position x of horse as a function of time t in radar system.
-The position of the horse at x = 2m in our coordinate system
Solution:
- The position of horse according to radar gun:
v(t) = dx / dt = 10 + 2*t^2
- Separate variables:
dx = (10 + 2*t^2).dt
- Integrate over interval x = 0 @ t= 0
x(t) = 10t + (2/3)*t^3
- time @ x = 2 :
2 = 10t + (2/3)*t^3
0 = 10t + (2/3)*t^3 + 2
- solve for t:
t = 0.1875 s
- Evaluate x* at t = 0.1875 s
x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3
x*(0.1875) = 10.18 m