Here's the procedure explained: Assume F represents the portion of the rope that is extending over the table. In this scenario, the frictional force that holds the rope on the table can be calculated using the formula: Ff = u*(1-f)*m*g. Additionally, it is important to determine the gravitational force that attempts to pull the rope off the table, Fg, calculated through: Fg = f*m*g. You then need to set these two equations equal to each other and resolve for f: f*m*g = u*(1-f)*m*g leads to f = u*(1-f) = u - uf. Simplifying gives f + uf = u, which results in f = u/(1+u) representing the fraction of the rope. This will lead you to the final answer.
Answer:
All three pendulums will have the same angular frequencies.
Explanation:
For a simple pendulum, the time period using the approximation 
is expressed as:
The angular frequency 
is defined as
Since the angular frequency remains unaffected by the initial angle (valid strictly for small angle approximations), we deduce that the angular frequencies of the three pendulums are identical.
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
Response:
A.3.13x10^14 electrons
B.330A/m²
C.9.11x10^5N/C
D. 0.23W
.Please review the attached document for further explanations
