Answer:
E = 1/2 M V^2 + 1/2 I ω^2 = 1/2 M V^2 + 1/2 I V^2 / R^2
E = 1/2 M V^2 (1 + I / (M R^2))
For a cylinder I = M R^2
For a sphere I = 2/3 M R^2
E(cylinder) = 1 + 1 = 2 omitting the 1/2 M V^2
E(sphere) = 1 + 2/3 = 1.67
E(cylinder) / E(sphere) = 2 / 1.67 = 1.2
The cylinder initially has 1.20 the energy of the sphere
The PE attained is proportional to the initial KE
H(sphere) = 2.87 / 1/2 = 2.40 m since it has less initial KE
Answer:
Explanation:
Two charged particles will vary in energy as we alter the distance between them.
Answer:
It's the third one in the picture ...
r = 3x - 2y
Explanation:
Answer:
<h2>a) 50°</h2><h2>b) 40°</h2>
Explanation:
Check the complete diagram n the attachment below
a) The angle of incidence on a plane surface is the angle between the incidence ray and the normal ray acting on a plane surface. The normal ray is the ray perpendicular to the surface while the incidence ray is the ray striking a plane surface.
According to the diagram, the angle of reflection r₂ on M₂ is 90°-g where g is the angle of glance.
Given angle of glance on M₂ to be 40°, r₂ = 90-40 = 50°
According the second law of reflection, the angle of incidence = angle of reflection, therefore i₂ = r₂ = 50° (on M₂)
Also ∠OO₂O₁ = ∠OO₁O₂ = 40° (angle of glance on M₁){alternate angle}
The angle of incidence on M₁ = 90° - 40° = 50°
b) The angle of incidence to the surface of M₁(∠PO₁A)will be the angle of glance on M₁ which is equivalent to 40°