For the velocity graph: start at 0s and 4m/s and draw a straight line to 2s and 2 m/s. Then draw a straight horizontal line to 4s and 2m/s
For the acceleration graph: start with a horizontal line from 0s and 2m/s/s to 2s and 2m/s/s. The draw another line from 2 s and 0m/s/s to 4 s and 0m/s/s
The answer is D) Velocity
Answer:
A) 138.8g
B)73.97 cm/s
Explanation:
K = 15.5 Kn/m
A = 7 cm
N = 37 oscillations
tn = 20 seconds
A) In harmonic motion, we know that;
ω² = k/m and m = k/ω²
Also, angular frequency (ω) = 2π/T
Now, T is the time it takes to complete one oscillation.
So from the question, we can calculate T as;
T = 22/37.
Thus ;
ω = 2π/(22/37) = 10.5672
So,mass of ball (m) = k/ω² = 15.5/10.5672² = 0.1388kg or 138.8g
B) In simple harmonic motion, velocity is given as;
v(t) = vmax Sin (ωt + Φ)
It is from the derivative of;
v(t) = -Aω Sin (ωt + Φ)
So comparing the two equations of v(t), we can see that ;
vmax = Aω
Vmax = 7 x 10.5672 = 73.97 cm/s
To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,
Q = V*A
Where,
A= Cross-sectional Area
V = Velocity
The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,


Our values are given as,


Re-arrange the equation to find the first ratio of rates we have:



The second ratio of rates is


