What class is that in if math or biology I’m not good that
<h2>
Answer: higher mean annual rainfall and temperatures. </h2>
Explanation:
Chemical weathering is the set of destructive processes through which rocky materials go trhough. These processes cause changes in the color, texture, composition, firmness and shape of the material.
It should be noted that this happens when the rocks come into contact with atmospheric agents such as oxygen and carbon dioxide.
Another important aspect is that rocks are able to break up more easily thanks to this type of weathering, since <u>the mineral grains within the rock lose adherence and dissolve better under the action of some physical agents</u>, such as <u>humidity (rainfall included) and temperature</u>.
Therefore:
Chemical weathering is greatest under conditions of <u>higher mean annual rainfall and temperatures. </u>
Change in position of object = Displacment
The wave speed completely depends on the characteristics and properties of the medium . . . physical properties for mechanical waves, electrical properties for electromagnedtic waves.
So if you want to change the speed of a wave, you have to change the medium . . . shoot it through some different kind of stuff. <em>(B) </em>
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>
