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
They are attractive
They don’t depend on charge
Answer:
0.12
Explanation:
The acceleration due to gravity of a planet with mass M and radius R is given as:
g = (G*M) / R²
Where G is gravitational constant.
The mass of the planet M = 3 times the mass of earth = 3 * 5.972 * 10^24 kg
The radius of the planet R = 5 times the radius of earth = 5 * 6.371 * 10^6 m
Therefore:
g(planet) = (6.67 * 10^(-11) * 3 * 5.972 * 10^24) / (5 * 6.371 * 10^6)²
g(planet) = 1.18 m/s²
Therefore ratio of acceleration due to gravity on the surface of the planet, g(planet) to acceleration due to gravity on the surface of the planet, g(earth) is:
g(planet)/g(earth) = 1.18/9.8 = 0.12
Answer:
a ) = 381.48 J
b )= 84.25 cm
Explanation:
Kinetic energy of the runner
= 1/2 m v²
= .5 x 66 x 3.4²
= 381.48 J
The final kinetic energy of the runner is zero .
Loss of mechanical energy
= 381.48 J
This loss in mechanical energy is due to action of frictional force .
b )
Let s be the distance of slide
deceleration due to frictional force
= μmg/m
.7 x 66 x 9.8 / 66
a = - 6.86 m s⁻¹
v² = u² - 2 a s
0 = 3.4² - 2x6.86 s
s = 3.4² / 2x6.86
= .8425 m
84.25 cm