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
This question is incomplete. The query is regarding a 3.00 cm diameter coin rolling up a 30.0° incline. With an initial angular speed of 60.0 rad/s, it rolls without slipping. Given that the moment of inertia of the coin is (1/2) MR², the distance the coin travels up the incline is calculated as 0.124 m.
Answer:
The radius is 
Explanation:
The problem states that
The magnetic field is 
The electron kinetic energy is 
In general, for a collision to happen, the centripetal force on the electron in its orbit must equal the magnetic force acting on it
This can be mathematically expressed as
=> 
Where m denotes the electron’s mass, which has a value of 
v signifies the escape velocity, mathematically represented as
Thus,
applying indices
substituting these values
