Answer:
The maximum velocity is 1.58 m/s.
Explanation:
A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.
Spring constant, K = 100 N/m
mass, m = 0.1 kg
Amplitude, A = 5 cm = 0.05 m
Let the angular frequency is w.
The maximum velocity is
Given: Change of x is 35.4m, Velocity Final=7.10 m/s, Velocity Initial=0m/s
Find: Acceleration
Analysis:
Vf²=Vi²+2aΔx (Velocity final squared equals Velocity initial squared plus 2 times acceleration times change of x)
(7.10 m²/s)²=(0 m/s)²+2a(35.4 m)
50.41 m/s²=(70.8 m)a
a=0.712 m/s²
Answer:
rpm
Explanation:
Given that rotational kinetic energy = 
Mass of the fly wheel (m) = 19.7 kg
Radius of the fly wheel (r) = 0.351 m
Moment of inertia (I) = 
Rotational K.E is illustrated as 
Since 1 rpm = 
Answer:
B = 0.546 T, F = 2.59 10⁻¹² N
Explanation:
The magnetic force is
F = q v x B
We can calculate the magnitude of the force and find the direction by the right hand rule
F = q v B sin θ
Let's use Newton's second law
F = m a
Acceleration is centripetal
a = v² / r
We substitute
q v B sin θ = m v² / r
The angle between the field and the radius of the circle is 90º so sin 90 = 1
q B = m v / r
B = m v / q r
Let's calculate ’
B = 1.67 10⁻²⁷ 2.97 10⁷ / (1.60 10⁻¹⁹ 0.568)
B = 0.546 T
The foce is
F = q v B
F = 1.60 10⁻¹⁹ 2.97 10⁷ 0.546
F = 2.59 10⁻¹² N
