Answer:
v = 4.4271 m/s
Explanation:
Given
m = 3 Kg
R = 0.2 m
∅ = 15°
h = 1.5 m
g = 9.8 m/s²
v = ?
Ignoring frictional losses, at the bottom of the plane
Total kinetic energy is = Potential Energy at the top of plane
Using Law of conservation of energy we have
U = Kt + Kr
m*g*h = 0.5*m*v² + 0.5*I*ω²
knowing that
Icylinder = 0.5*m*R²
ω = v/R
we have
m*g*h = 0.5*m*v² + 0.5*(0.5*mR²)*(v/R)² = 0.75*m*v²
⇒ v = √(g*h/0.75) = √(9.8 m/s²*1.5 m/0.75)
⇒ v = 4.4271 m/s
Well its C, cant show you the work its in my head sorry.
Answer: 1.04N
Explanation:
Given
q1 = 2*10^-6C
q2 = 3.6*10^-6C
r = 0.25m
k = 9*10^9
Magnitude of electrostatic force can be calculated by using coulomb's law. Coulomb's law states that, "the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them."
F =(kq1q2) / r²
F = (9*10^9 * 2*10^-6 * 3.6*10^-6) / 0.25²
F = 0.0648/0.0625
F = 1.04N
The type of electrostatic force between the charges is the repulsive force
