a man pulls a box with a force of 90.00 N directed along the handle which makes angle of 45 degrees with the ground. Find the ho
1 answer:
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At a distance of 469.2 m from the original point below the airplane.
Explanation:
First of all, we have to calculate the time it takes for the sack to reach the ground.
To do so, we just analyze its vertical motion, which is a free-fall motion, so we can use the suvat equation:
where, taking downward as positive direction:
s = 300 m is the vertical displacement
u = 0 is the initial vertical velocity
t is the time
is the acceleration of gravity
Solving for t, we find the it takes for the sack to reach the ground:
Now we analyze the horizontal motion. The horizontal velocity of the pack is constant (since there are no forces along the horizontal direction) and equal to the initial speed of the airplane, so:
We also know the total time of flight,
t = 7.82 s
Therefore, we can find the horizontal distance travelled by the sack:
So, the sack will land 469.2 m from the original point below the airplane.
Learn more about free fall and projectile motion:
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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