We are given an object that is speeding up on a level ground.
Let's remember that the gravitational energy depends on the change in height, therefore, if the object is not changing its height it means that the gravitational energy remains constant.
The kinetic energy depends on the velocity. If the velocity is increasing this means that the kinetic energy is also increasing.
Now, every change in velocity requires acceleration and acceleration requires a force. The force and the distance that the object moves are equivalent to the work that is transferred to the object and therefore, the change in kinetic energy. This means that the total energy of the system increases as work is transferred to the mass.
We have that the total energy of the system increases in the form of kinetic energy and that the gravitational potential energy remains constant. Therefore, the diagrams should look like pie charts that grow but the area of the segment of the potential energy stays the same. It should look similar to the following.
<span>The intensity of an earthquake is dependent on one's proximity to the focus of the quake, also called the "epicenter" and is based on observations of the shaking of the ground on humans, structures, and the natural landscape.</span>
Answer:
12.0 meters
Explanation:
Given:
v₀ = 0 m/s
a₁ = 0.281 m/s²
t₁ = 5.44 s
a₂ = 1.43 m/s²
t₂ = 2.42 s
Find: x
First, find the velocity reached at the end of the first acceleration.
v = at + v₀
v = (0.281 m/s²) (5.44 s) + 0 m/s
v = 1.53 m/s
Next, find the position reached at the end of the first acceleration.
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²
x = 4.16 m
Finally, find the position reached at the end of the second acceleration.
x = x₀ + v₀ t + ½ at²
x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²
x = 12.0 m
Ng seismic and translational waves we get the law of michio kaku.
