Newton's second law allows calculating the response for the person's acceleration while leaving the trampoline is:
-4.8 m / s²
Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body
F = m a
Where the bold letters indicate vectors, F is the force, m the masses and the acceleration
The free body diagram is a diagram of the forces without the details of the body, in the attached we can see the free body diagram for this system
-W = m a
Whera
is the trampoline force
Body weight is
W = mg
We substitute
- mg = ma
a =
Let's calculate
a = 
a = -4.8 m / s²
The negative sign indicates that the acceleration is directed downward.
In conclusion using Newton's second law we can calculate the acceleration of the person while leaving the trampoline is
-4.8 m / s²
Learn more here: brainly.com/question/19860811
Explanation:
It is given that,
Fundamental frequency, f = 220 Hz
(a) We know that at 0 degrees, the speed of sound in air is 331 m/s.
For open pipe, 
l is the length of pipe
Also,


(b) Let f' is the fundamental frequency of the pipe at 30 degrees and v' is its speed.


v' = 348.71 m/s
So, 

f' = 232.4 Hz
Hence, this is the required solution.
Answer:
The Position of the object L = 0.172 m
Explanation:
The detailed explanation of the question is given in the attach document.
Answer : The force will be 4501.9
We can see that, two forces acting on the dummy in two different direction.
We know that, here two forces are given in perpendicular direction with each other.
We know the force is the vector addition law so, we will use the Pythagoras theorem for the resultant of the vectors
Now, the net force is

Two forces is given,


Now, the net force is



Hence, the force will 4501.9 N