Answer:
the distance from the location of the center of gravity to the location of the center o mass for this system is 1.13m
Explanation:
Given that
m₁=4.4kg
x₁=+1.1m
m₂=3.7kg
x₂=+0.80m
m₃=2.9kg
x₃=+1.6m
The position of the center of mass is
Xcm = [m₁x₁ +m₂x₂ +m₃x₃]/(m₁+m₂+m₃)
= [(4.40kg)(1.1 m)+(3.70 kg)(0.80 m)+(2.90 kg)(1.60 m)]/(4.4 kg + 3.70 kg+2.90 kg)
= 1.13 m
The position of the center of gravity is 1.13m
Therefore, the distance from the location of the center of gravity to the location of the center o mass for this system is 1.13m
Explanation:
When the wires of a circuit will break, circuit will be open state and stop the flow of current. Once the wires are reconnected, current will start flowing again in the circuit.
Answer:
k = 1400.4 N / m
Explanation:
When the springs are oscillating a simple harmonic motion is created where the angular velocity is
w² = k / m
w =
where angular velocity, frequency and period are related
w = 2π f = 2π / T
we substitute
2π / T = \sqrt{ \frac{k}{m} }
T² = 4π²
k = π²
in this case the period is T = 1.14s, the combined mass of the children is
m = 92.2 kg and the constant of the two springs is
k = 4π² 92.2 / 1.14²
k = 2800.8 N / m
to find the constant of each spring let's use the equilibrium condition
F₁ + F₂ - W = 0
k x + k x = W
indicate that the compression of the two springs is the same, so we could replace these subtraction by another with an equivalent cosecant
(k + k) x = W
2k x = W
k_eq = 2k
k = k_eq / 2
k = 2800.8 / 2
k = 1400.4 N / m