Answer:
35700000
Explanation:
multiply the meter value by 1e+9
M = 2.2 g = 2.2 x 10⁻³ kg, the mass of the bug.
r = 3.0 cm = 0.03 m, the radial distance from the center.
The angular speed is
ω = 280 rpm
= (280 rev/min)*(2π rad/rev)*(1/60 min/s)
= 29.3215 rad/s
The moment of inertia of the bug is
I = mr²
= (2.2 x 10⁻³ kg)*(0.03 m)²
= 1.98 x 10⁻⁶ kg-m²
Calculate the angular momentum of the bug.
J = Iω
= (1.98 x 10⁻⁶ kg-m²)*(29.3215 rad/s)
= 5.806 x 10⁻⁵ (kg-m²)/s
Answer: 5.806 x 10⁻⁵ (kg-m²)/s
Answer:462400joules
Explanation:
Mass=m=800kg velocity=v=34m/s
Kinetic energy=(m x v^2)/2
Kinetic energy =(800x34^2)/2
Kinetic energy =(800x34x34)/2
kinetic energy =924800/2
Kinetic energy =462400joules
Mass of the block = 1.4 kg
Weight of the block = mg = 1.4 × 9.8 = 13.72 N
Normal force from the surface (N) = 13.72 N
Acceleration = 1.25 m/s^2
Let the coefficient of kinetic friction be μ
Friction force = μN
F(net) = ma
μmg = ma
μg = a
μ = 
μ = 
μ = 0.1275
Hence, the coefficient of kinetic friction is: μ = 0.1275
