You are removing branches from your roof after a big storm. You throw a branch horizontally from your roof, which is a height 3.
1 answer:
Answer:
The initial velocity in the x-direction with which the branch was thrown is approximately 10.224 m/s
Explanation:
The given parameters of the motion of the branch are;
The height from which the branch is thrown = 3.00 m
The horizontal distance the branch lands from where it was thrown, x = 8.00 m
The direction in which the branch is thrown = Horizontally
Therefore, the initial vertical velocity of the branch, = 0 m/s
The time it takes an object in free fall (zero initial downward vertical velocity) to reach the ground is given as follows;
s = ·t + 1/2·g·t²
Where;
= 0 m/s
s = The initial height of the object = 3.00 m
g = The acceleration due to gravity = 9.8 m/s²
∴ s = 0·t + 1/2·g·t² = 0 × t + 1/2·g·t² = 1/2·g·t²
t = √(2·s/g) = √(2 × 3/9.8) = (√30)/7 ≈ 0.78246
The horizontal distance covered before the branch touches the ground, x = 8.00 m
Therefore, the initial velocity in the horizontal, x-direction with which the branch was thrown, 'uₓ', is given as follows;
uₓ = x/t = 8.00 m/((√30)/7 s)
Using a graphing calculator, we get;
uₓ = 8.00 m/((√30)/7 s) = (28/15)·√30 m/s ≈ 10.224 m/s
The initial velocity in the horizontal, x-direction with which the branch was thrown, uₓ ≈ 10.224 m/s.
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Answer:
v = -1.8t+36
20 seconds
360 m
40 seconds
36 m/s
The object speed will increase when it is coming down from its highest height.
Explanation:
Differentiating with respect to time we get
a) Velocity of the object after t seconds is v = -1.8t+36
At the highest point v will be 0
b) The object will reach the highest point after 20 seconds
c) Highest point the object will reach is 360 m
d) Time taken to strike the ground would be 20+20 = 40 seconds
Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of
e) The velocity with which the object strikes the ground will be 36 m/s
f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.
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