Answer:
Explanation:
The angular momentum of an object is given by:
where
m is the mass of the object
v is its velocity
r is the distance of the object from axis of rotation
Here we have:
m = 350 g = 0.35 kg is the mass of the ball
v = 9.0 m/s is the velocity
r = 3.0 m is the distance of the object from axis of rotation (if we take the ground as the centre of rotation)
Therefore, the angular momentum is:
Answer:
C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂
Explanation:
You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.
Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.
Stone 1
y₁ = v₀₁ t + ½ g t²
y₁ = 0 + ½ g t²
Rock2
It comes out a little later, let's say a second later, we can use the same stopwatch
t ’= (t-t₀)
y₂ = v₀₂ t ’+ ½ g t’²
y₂ = 0 + ½ g (t-t₀)²
y₂ = + ½ g (t-t₀)²
Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to
S = y₁ -y₂
S = ½ g t²– ½ g (t-t₀)²
S = ½ g [t² - (t²- 2 t to + to²)]
S = ½ g (2 t t₀ - t₀²)
S = ½ g t₀ (2 t -t₀)
This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.
For t <to. The rock y has not left and the distance increases
For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time
passes
Now we can analyze the different statements
A) false. The difference in height increases over time
B) False S increases
C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂
Answer:
475
Explanation:
Cori does not exert any more force than 475 J, so 475 is the answer.
Answer:
L = 5076.5 kg m² / s
Explanation:
The angular momentum of a particle is given by
L = r xp
L = r m v sin θ
the bold are vectors, where the angle is between the position vector and the velocity, in this case it is 90º therefore the sine is 1
as we have two bodies
L = 2 r m v
let's find the distance from the center of mass, let's place a reference frame on one of the masses
= 
i
x_{cm} = 
x_{cm} = 
x_{cm} = 
x_{cm} = 13.1 / 2 = 6.05 m
let's calculate
L = 2 6.05 74.3 5.65
L = 5076.5 kg m² / s
