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
Answer:
The value of the angle theta = 135degree, which is option c.
Explanation :
These are the options given in the question
a) 45 degrees
b)120 degrees
c)135 degrees
d)150 degrees
e) it is impossible to deliver a negative power.
The concept of work done on an object is applied here.
work = Fd
F = ma
work = mad
Work is a form of energy transferred by an object after the application of a force that act on the object. A step by step explanation is as attached below.
By definition, the density of an object is given by:

Where,
M: mass of the object
V: volume of the object
Since the mass and volume of an object are numerical values greater than zero, then it follows that:

It is important to respect the units of each measure.
For this case we can use the grams for the mass and cubic centimeters for the volume.
Answer:
A possible value for density is given by:

Answer: The minimum time between pulses (in fresh water)
= 0.3106 s
Explanation:
To calculate the speed of echoes sounds, we will use
Speed = 2x/t
Where x = distance and t = total time.
Please note that:
sound travels at 343 m/s in air. But it travels at 1,481 m/s in water (almost 4.3 times as fast as in air);
Total distance = 2 × 230 = 460m
Using speed of sound in water
1481 = 460/t
t = 460/1481
t = 0.3106