It you balance a weight of 20 N at the 15 cm MARK and want to balance a weight of on the other side of the meter stick at 70 cm
1 answer:
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Answer:
f = 5.3 Hz
Explanation:
To solve this problem, let's find the equation that describes the process, using Newton's second law
∑ F = ma
where the acceleration is
a =
B- W = m \frac{d^2 y}{dt^2 }
To solve this problem we create a change in the reference system, we place the zero at the equilibrium point
B = W
In this frame of reference, the variable y' when it is oscillating is positive and negative, therefore Newton's equation remains
B’= m
the thrust is given by the Archimedes relation
B = ρ_liquid g V_liquid
the volume is
V = π r² y'
we substitute
- ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }
this differential equation has a solution of type
y = A cos (wt + Ф)
where
w² = ρ_liquid g π r² /m
angular velocity and frequency are related
w = 2π f
we substitute
4π² f² = ρ_liquid g π r² / m
f =
calculate
f =
f = 5.3 Hz
Answer:
The displacement was 320 meters.
Explanation:
Assuming projectile motion and zero initial speed (i.e., the object was dropped, not thrown down), you can calculate the displacement using the kinematic equation:
The displacement was 320 meters.