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Based on the data provided, the impulse of the floor on the ball is 59.4 Ns.
<h3>What is the impulse of the floor on the ball?</h3>
Using the equation of motion to determine the velocity at the end of the fall
Where v is velocity at the end of fall
u is initial velocity = 0
g is acceleration due to gravity = 9.81 m/s^2
h is height = 20
- Taking downward velocity as negative and up as positive
v^2 = 0 + 2 (9.81)(20)
v^2 = 392.4
v = - 19.8 m/s
The velocity, v after bouncing is calculated also:
u = 0
g = 9.81 m/s^2
h = 5.0 m
v^2 = 0 + 2(9.81)(5)
v^2 = 98.1
v = 9.904 m/s
- Impulse = change in momentum
- Impulse = m(v- u)
Impulse = 2.0 × (9.9 -(-19.8)
Impulse = 59.4 Ns
Therefore, the impulse of the floor on the ball is 59.4 Ns.
Learn more about impulse at: brainly.com/question/904448
When two waves (either mechanical or electromagnetic) combine in such a way that their peaks and troughs combine to produce a wave of large amplitude, the wave behaviour is known as constructive interference. The opposite process is where the peaks of one wave combines with the troughs of another wave in order to produce a wave of smaller amplitude. This process is known as destructive interference.