Answer:
Tectonic plate interactions are of three different basic types: Divergent boundaries are areas where plates move away from each other, forming either mid-oceanic ridges or rift valleys. These are also known as constructive boundaries. Convergent boundaries are areas where plates move toward each other and collide.
Explanation:
Meaning the answer to your question is depending on what type of tectonic plate interaction is occurring will depend on how the plates interact.
Answer
Explanation:
The question was incomplete as the events are not given in the question. However the answer to your question is given as follows. The correct order of the events from youngest (top) to oldest (bottom) is given as follows.
Moon formation
↑
Earth formation
↑
Nuclear fusion in protosun
↑
BigBang
Answer:
a) During the reaction time, the car travels 21 m
b) After applying the brake, the car travels 48 m before coming to stop
Explanation:
The equation for the position of a straight movement with variable speed is as follows:
x = x0 + v0 t + 1/2 a t²
where
x: position at time t
v0: initial speed
a: acceleration
t: time
When the speed is constant (as before applying the brake), the equation would be:
x = x0 + v t
a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:
x = 0m + 26 m/s * 0.80 s = <u>21 m </u>
b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):
v = v0 + a* t
0 = 26 m/s + (-7.0 m/s²) * t
-26 m/s / - 7.0 m/s² = t
t = 3.7 s
With this time, we can calculate how far the car traveled during the deacceleration.
x = x0 +v0 t + 1/2 a t²
x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>
Answer:
Vf = 15 m/s
Explanation:
First we consider the upward motion of ball to find the height reached by the ball. Using 3rd equation of motion:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity = -9.8 m/s² (negative sign for upward motion)
h = height =?
Vf = Final Velocity = 0 m/s (Since, ball momentarily stops at highest point)
Vi = Initial Velocity = 15 m/s
Therefore,
2(-9.8 m/s²)h = (0 m/s)² - (15 m/s)²
h = (-225 m²/s²)/(-19.6 m/s²)
h = 11.47 m
Now, we consider downward motion:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity = 9.8 m/s²
h = height = 11.47 m
Vf = Final Velocity = ?
Vi = Initial Velocity = 0 m/s
Therefore,
2(9.8 m/s²)(11.47 m) = Vf² - (0 m/s)²
Vf = √(224.812 m²/s²)
<u>Vf = 15 m/s</u>
