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 answer is:
V = d/t d = 86 km t = 1.3 hrs
V = 86 km/ 1.3 hrs
V = 66.15 km/ hrs
I hope this helps!!
Formula for terminal
velocity is:
Vt = √(2mg/ρACd)
<span>Vt = terminal velocity = ?
<span>m = mass of the falling object = 72 kg
<span>g = gravitational acceleration = 9.81 m/s^2
<span>Cd = drag coefficient = 0.80
<span>ρ = density of the fluid/gas = 1.2 kg/m^3</span>
<span>A = projected area of the object (feet first) = 0.21 m * 0.41
m = 0.0861 m^2
Therefore:</span></span></span></span></span>
Vt = √(2 * 72
* 9.81 / 1.2 * 0.0861 * 0.80)
<span>Vt = 130.73 m/s</span>
The west constituent of their sequence needs to cancel out 58 mph crosswind. Subsequently a northwest direction is a 45-degree angle up to even with the destination. That is the third point out of the triangle and the right angle is at the destination. The top side is the west constituent of their flight the vertical side is their resultant travel and the hypotenuse is their definite distance flown. Since the 58 mph crosswind was negated by flying northwest, the distance from the beginning to the destination must be the same distance as the west component of their travel. The hypotenuse is square root of twice the side since it has 2 identical sides.
c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02
Alternative solution:
c = sqrt (2) * 58 = 1.414 * 58 = 82.02
Therefore, they have to fly 82.02 mph
