We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.
1) Equilibrium of forces:
where
is the weight of the person
is the weight of the scaffold
Re-arranging, we can write the equation as
(1)
2) Equilibrium of torques:
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using
and replacing T1 with (1), we find
from which we find
And then, substituting T2 into (1), we find
|momentum| = (mass) x (speed)
Momentum of each arrow = (0.1 kg) x (30 m/s) = 3 kg-m/s
Everything is equal for the two arrows, except their direction.
So the total momentum vector of the 2-arrow system consists of
-- a 3 kg-m/s component pointing east
plus
-- a 3 kg-m/s component pointing south.
The magnitude of the total momentum is
√ [ (3 kg-m/s)² + (3 kg-m/s)² ]
= √ [ 9 kg²-m²/s² + 9 kg²-m²/s² ]
= √ [ 18 kg²-m²/s² ]
= 3√2 kg-m/s (approx 4.243 kg-m/s)
The direction of the total momentum vector is the angle
south of east whose tangent is
(3 kg-m/s) / (3 kg-m/s)
= tan⁻¹ (1)
= 45° south of east.
That direction is southeast.
Answer:
when two electrons are in close proximity Each atom's electron begins to detect the proton of the other atom. As a result, it is attracted to not just its own proton, but also the proton of the other atom.
Answer:
i think the answer is A but i am not sure
Explanation: