Answer:

ΔK = 2.45 J
Explanation:
a) Using the law of the conservation of the linear momentum:

Where:


Now:

Where
is the mass of the car,
is the initial velocity of the car,
is the mass of train,
is the final velocity of the car and
is the final velocity of the train.
Replacing data:

Solving for
:

Changed to cm/s, we get:

b) The kinetic energy K is calculated as:
K = 
where M is the mass and V is the velocity.
So, the initial K is:



And the final K is:




Finally, the change in the total kinetic energy is:
ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J
So I'm a junior. I am currently taking AP Calc BC and AP Physics B.
As of now, I'm not sure if I should take AP Probability and Statistics or Differential Equations/Calc III next year. Also, I'm debating between taking AP Physics C or AP Chemistry.
Which ones do you think would look better on a transcript? I heard that Diffeq/CalcIII is harder than AP ProbStat, but ProbStat is an AP course which will be weighted heavier. Also, should I take Physics C since i've taken Physics B this year already?
Acceleration is the rate of change of velocity
To define acceleration, We need to know more about motion.
Motion: This can be defined as the change in position of a body from one point to another. When an object accelerates, it undergoes motion.
<u>Definition</u>
Acceleration can be defined as the rate of change of velocity. The S.I unit of acceleration is meter-per-squared seconds. (m/s²)
The formula of acceleration is
- a = (v-u)/t................. Equation 1
⇒ Where:
- a = acceleration
- u = initial velocity
- v = final velocity
- t = time
Hence, Acceleration is the rate of change of velocity
Learn more about acceleration here: brainly.com/question/605631
Answer: 1018.26 m/s
Explanation:
Approaching the orbit of the Moon around the Earth to a circular orbit (or circular path), we can use the equation of the speed of an object with uniform circular motion:
Where:
is the speed of travel of the Moon around the Earth
is the Gravitational Constant
is the mass of the Earth
is the distance from the center of the Earth to the center of the Moon
Solving:
This is the speed of travel of the Moon around the Earth
I believe this would be an example of Mary's velocity. We have her speed and direction which is all you need to find velocity.