Potential energy (PE ) = m g h
Where:
m = mass = 3800 kg
g = acceleration due gravity = 10 m/s^2
h = heigth = 110 meters
Replacing:
PE = 3800 * 10 * 110 = 4,180,000 J
3. is the answer, <span>Sodium needs to lose one electron, and chlorine needs to gain one electron. This is because Sodium's row always wants to give away an electron, while Chlorine's row wants to gain an electron.</span>
Answer:
A physical change involves a change in physical properties. Examples of physical properties include melting, transition to a gas, change of strength, change of durability, changes to crystal form, textural change, shape, size, color, volume and density.
Answer:

Explanation:
<u>Accelerated Motion
</u>
When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

where a is the acceleration, and vo is the initial speed
.
The train has two different types of motion. It first starts from rest and has a constant acceleration of
for 182 seconds. Then it brakes with a constant acceleration of
until it comes to a stop. We need to find the total distance traveled.
The equation for the distance is

Our data is

Let's compute the first distance X1


Now, we find the speed at the end of the first period of time


That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0



Computing the second distance


The total distance is


