A skier (down a straight course) is able to enter the starting gate with a speed of 1 m/sec and the average acceleration down th
e hill was 3 m/s2. If
the skier covers the course in 10 sec, how fast was he moving at the bottom of the course?
O31 m
O 13 m/s
O-29 m/s
O 31 m/s
the skier covers the course in 10 sec, how fast was he moving at the bottom of the course?
O31 m
O 13 m/s
O-29 m/s
O 31 m/s
1 answer:
You might be interested in
According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.
To solve this problem we must apply the concept related to the conservation of energy theorem.
PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so
PART B) Replacing the values given as,
Therefore the speed of the masses would be 1.8486m/s
Answer:
a) I= I₀ (cos²θ - cos⁴θ) b) 75.5º
Explanation:
a) For this exercise we must use Malus's law
I = I₀ cos² θ
where tea is the angle between the two polarizers.
We apply this expression to our case
* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical
I₁ = I₀ cos² θ
* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.
The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical
θ₂ = 90- θ’ = θ
fiate that in the exercise we must take a reference system and measure everything with respect to this system.
I = I₁ cos² θ'
we substitute
I = (I₀ cos² tea) cos² (θ - 90)
cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ
I = Io cos² θ sin² θ
1= cos²θ+ sin²θ
sin²θ = 1 - cos²θ
I= I₀ (cos²θ - cos⁴θ)
b) to find when the intensity is maximum,
we can use that we have an extreme point when the drift is zero
= 0
\frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0
whereby
cos θ - 2 cos³ θ = 0
cos θ ( 1 - 2 cos² θ) = 0
The zeros of this function are in
θ = 90º
1-2cos²θ =0 cos θ = 0.25 θ = 75.5º
Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.
Consequently, the angle that allows the maximum intensity to pass is 75.5º