A water skier is pulled behind a boat. When the skis push down on the water, what is the reaction force?
2 answers:
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Answer:
B) 27.3 m
Explanation:
The rock describes a parabolic path.
The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).
The equation of uniform rectilinear motion (horizontal ) for the x axis is :
x = vx*t Equation (1)
Where:
x: horizontal position in meters (m)
t : time (s)
vx: horizontal velocity in m/s
The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:
(vfy)² = (v₀y)² - 2g(y- y₀) Equation (2)
vfy = v₀y -gt Equation (3)
Where:
y: vertical position in meters (m)
y₀ : initial vertical position in meters (m)
t : time in seconds (s)
v₀y: initial vertical velocity in m/s
vfy: final vertical velocity in m/s
g: acceleration due to gravity in m/s²
Data
v₀ = 30 m/s , at an angle α=40.0° above the horizontal
v₀x = vx = 30*cos40° = 22.98 m/s
v₀y = 30*sin40° = 19.28 m/s
y₀ = 2m
y = 18.0 m
g = 9.8 m/s²
Calculation of the time (t) it takes for the rock to reach at 18 m above the ground
We replace data in the equation (2)
(vfy)² = (v₀y)² - 2g(y- y₀)
(vfy)² = (19.28)² - 2(9.8)(18- 2)
(vfy)² = 371.86 - 313.6
(vfy)² = 58.26
vfy = 7.63 m/s
We replace vfy = 7.63 m/s in the equation (2)
vfy = v₀y - gt
7.63 = 19.28 - (9.8)(t)
(9.8)(t) = 11.65
t = 11.65 / (9.8)
t = 1.19 s
Horizontal distance from where the rock was thrown to the window
We replace t = 1.19 s , in the equation (1)
x = vx*t
x = (22.98)* ( 1.19 )
x = 27.3 m
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