A force of 20N pushes an object of mass 5.0kg along a rough surface of 5.0N
2 answers:
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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
The kinetic energy of the electron is
where
is the mass of the electron and v its speed. Since we know the value of the kinetic energy, 
, we can find the value of the speed v:
where
Answer:
Part a)
Part b)
Part c)
Part d)
Part e)
Part f)
Part g)
Explanation:
Initial speed of the launch is given as
initial speed =
angle = degree
Now the two components of the velocity
similarly we have
Part a)
Now we know that horizontal range is given as
maximum height is given as
so we have
time of flight is given as
Part b)
Now the speed of the ball in x direction is always constant
so at the peak of its path the speed of the ball is given as
Part c)
Initial vertical velocity is given as
Part d)
Initial speed is given as
so we will have
Part e)
Angle of projection is given as
Part f)
If we throw at same speed so that it reach maximum height
then the height will be given as
Part g)
For maximum range the angle should be 45 degree
so maximum range is