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?
The acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².
<h3>How to calculate acceleration?</h3>
The acceleration of a freight train can be calculated using the following formula:
Force = mass × acceleration
According to this question, a 600,000kg freight train can produce 100,000N of force. The acceleration is as follows:
100,000 = 600,000 × a
100,000 = 600,000a
a = 0.167m/s²
Therefore, the acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².
Learn more about acceleration at: brainly.com/question/12550364
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Answer:
Explanation:
A pure substance is something that is entirely made up of particles that are identical to each other.
Any substance that is not pure, must be a mixture.
We are surrounded by mixtures. The air is a mixture of gases . The oceans are a mixture of (mainly) water and salt. The solid earth is mostly rock, which is a mixture of different minerals.
Natural resources are substances we need and use, which occur naturally. Some come from living things, (example) cotton other are non-living (example) sand.
The opposite of a natural resource is a made resource.
You draw a straight line from the start point to the end point. It doesn't matter what route was actually followed for the trip.
Answer:
a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g
Explanation:
For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.
The velocity of each ball is
ball 1. thrown upwards vo is positive
v² = v₀² - 2 g (y-y₀)
in this case the height y is zero and the height i = h
v = √(v₀² + 2g h)
ball 2 thrown down, in this case vo is negative
v = √(v₀² + 2g h)
The times to get to the ground
ball 1
v = v₀ - g t₁
t₁ = 
ball 2
v = -v₀ - g t₂
t₂ = - \frac{v_{o} + v }{ g}
From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is
Δt = t₂ -t₁
Δt = ![\frac{1}{g} \ [(v_{o} - v) - ( - v_{o} - v) ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bg%7D%20%5C%20%5B%28v_%7Bo%7D%20-%20v%29%20%20-%20%28%20-%20v_%7Bo%7D%20%20-%20v%29%20%5D)
Δt = 2 v₀ / g
