Hello there!
Physical changes do not change what a substance is, whereas a chemical change results in a new substance being formed, so they do change what a substance is.
I hope I could help you to solve this question. Have a great rest of your day! :)
Answer:
Explanation:
Torque, 
is given by
where F is force and r is perpendicular distance
where 
is the angle of inclination
Torque, can also be found by
where I is moment of inertia and a is angular acceleration
Therefore, Fr=Ia and F=mg where m is mass and g is acceleration due to gravity
Making a the subject, 
and already I is given as
hence
Taking g as 9.81, is given as 37 and L is 1.2
Plate tectonics<span>is a </span>scientific theory<span> that describes the large-scale motion of </span>Earth<span>'s </span>lithosphere<span>. This theoretical model builds on the concept of </span>continental drift<span> which was developed during the first few decades of the 20th century. The </span>geoscientific<span> community accepted plate-tectonic theory after </span>seafloor spreading was validated in the late 1950s and early 1960s.<span>The lithosphere, which is the rigid outermost shell of a planet (the crust and upper mantle), is broken up into </span>tectonic plates<span>. </span>
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>
