The hypothesis because its very hard to make and it confounds me
Answer:
a) L=0. b) L = 262 k ^ Kg m²/s and c) L = 1020.7 k^ kg m²/s
Explanation:
It is angular momentum given by
L = r x p
Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum
One of the easiest ways to make this vector product is with the use of determinants
![{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]](https://tex.z-dn.net/?f=%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%26z%5C%5Cpx%26py%26pz%5Cend%7Barray%7D%5Cright%5D)
Let's apply this relationship to our case
Let's start by breaking down the speed
v₀ₓ = v₀ cosn 45
voy =v₀ sin 45
v₀ₓ = 9 cos 45
voy = 9 without 45
v₀ₓ = 6.36 m / s
voy = 6.36 m / s
a) at launch point r = 0 whereby L = 0
. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero
vfy² = voy²- 2 g y
y = voy² / 2g
y = (6.36)²/2 9.8
y = 2.06 m
Let's calculate the angular momentum
L= ![\left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%260%5C%5Cpx%260%260%5Cend%7Barray%7D%5Cright%5D)
L = -px y k ^
L = - (m vox) (2.06) k ^
L = - 20 6.36 2.06 k ^
L = 262 k ^ Kg m² / s
The angular momentum is on the z axis
c) At the point of impact, at this point the height is zero and the position on the x-axis is the range
R = vo² sin 2θ / g
R = 9² sin (2 45) /9.8
R = 8.26 m
L =
L = - x py k ^
L = - x m voy
L = - 8.26 20 6.36 k ^
L = 1020.7 k^ kg m² /s
If it takes
![t](https://tex.z-dn.net/?f=t)
seconds to reach the car, then the distance
![d](https://tex.z-dn.net/?f=d)
is
![3.9t](https://tex.z-dn.net/?f=3.9t)
.
The bear's distance from the tourist's starting point is
![6t-23](https://tex.z-dn.net/?f=6t-23)
For maximum
![d](https://tex.z-dn.net/?f=d)
, we set the equations equal to each other:
![3.9t=6t-23](https://tex.z-dn.net/?f=3.9t%3D6t-23)
![\Rightarrow -2.1t=-23](https://tex.z-dn.net/?f=%5CRightarrow%20-2.1t%3D-23)
![\Rightarrow t=\frac{23}{2.1}](https://tex.z-dn.net/?f=%5CRightarrow%20t%3D%5Cfrac%7B23%7D%7B2.1%7D)
so the distance is
Friction as it will move charge (electrons) from one object to another