Answer:
b = 0.6487 kg / s
Explanation:
In the context of oscillatory motion, friction is related to velocity,
fr = - b v
where b represents the friction coefficient.
Upon solving the equation, the angular velocity is represented as
w² = k / m - (b / 2m)²
In this case, we're given an angular frequency w = 1Hz, the mass m = 0.1 kg, and the spring constant k = 5 N / m. This allows us to derive the friction coefficient.
Let’s denote
w₀² = k / m
w² = w₀² - b² / 4m²
b² = (w₀² -w²) 4 m²
Now, let's calculate the angular frequencies.
w₀² = 5 / 0.1
w₀² = 50
w = 2π f
w = 2π 1
w = 6.2832 rad / s
Substituting values yields
b² = (50 - 6.2832²) 4 0.1²
b = √ 0.42086
b = 0.6487 kg / s
Answer:
d_total = 12 m
Explanation:
In this kinematics scenario illustrated in the graph provided, we determine the distance traveled over a 24-second duration.
The comprehensive distance can be calculated as follows:
d_total = d₁ + d₂ + d₃
Given that d₂ on the graph is level (v=0), its distance equates to zero, hence d₂ = 0.
The distance for d₁ is calculated as:
d₁ = 12 - 6 = 6 m
For distance d₃:
d₃ = 6 - 0 = 6 m
Thus, the overall distance covered is:
d_total = 6 + 0 + 6
d_total = 12 m
Answer:
Explanation:
The four wires are arranged in series: this setup indicates that the same current passes through them, while the total voltage from the battery, V0, equals the combined voltages across each resistor:
Additionally, the total resistance for this series configuration is
Using Ohm's law, we can determine the voltage V2 across wire 2:
(1)
Here, I represents the entire current flowing through the circuit, which is calculated as:
By substituting into equation (1), we derive V2:
Answer:
1.5 × 10³⁶ light-years
Explanation:
A particular square area in interstellar space measures roughly 2.4 × 10⁷² (light-years)². To find the area of a square, the following formula is utilized:
A = l²
where,
A represents the area of the square
l denotes the length of one side of the square
Thus, l = √A = √2.4 × 10⁷² (light-years)² = 1.5 × 10³⁶ light-years
Given the values:
m = 1160 kg
g = 9.81 m/s²
v = 2.5 m/s
Unknowns include:
k (spring constant)
x (spring compression)
1. To address this, use Newton's second law and Hook's law:
2. The total energy in the spring must correspond to the satellite's energy:
By combining these two equations
