The locations of earthquakes support the theory of plate tectononics by the recorded happening are mostly on the outline of tectonic plates.
Answer:
Explanation:
When the object fall from the top floor it moves with parabolic motion.
This particular case is a horizontal motion of a projectile, so the equation that relates the distance and the time in the y-axes is:
Here: y is the final height, it will be 0, because the object touch the ground.
y₀ is the initial height, it will be h and v₀ in y-axis is zero, because it has a horizontal initial velocity.
So we have:
(1)
We know that the velocity at the x-axes remains constant, then we have:
x is the distance from the base of the building
we can solve it for t and put on (1).
Solving it for x we will have:
But
I hope it helps you!
To provide a greater certainty that the observed results are not by chance.
Answer:
Time period becomes one third
Explanation:
Time period is the reciprocal of frequency
so if frequency is tripled period will be one third.
Let u = the vertical launch speed.
Neglect air resistance, and g = 9.8 m/s².
At the maximum height of h = 120 cm = 1.2 m, the vertical velocity is zero.
Therefore
(u m/s)² - 2*(9.8 m/s²)*(1.2 m) = 0
u² = 23.52 (m/s)²
u = 4.85 m/s
The time to attain maximum height is one half of the travel time.
If t = time to attain maximum height, then
u - gt = 0
4.85 - 9.8t = 0
t = 4.85/9.8 = 0.495 s
Therefore the travel time is 2*0.0495 = 0.99 s
Answers:
(a) 4.85 m/s
(b) 0.99 s