A 0.500 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 40.0 pC charge on it
1 answer:
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Answer:
The magnitude of the electrostatic force decreases by a factor 9
Explanation:
The electrostatic force between two charges is given by:
where
k is the Coulomb's constant
q1 and q2 are the two charges
d is the distance between the two charges
We see that the magnitude of the force is inversely proportional to the square of the distance. If the distance is increased to 3d: d' = 3d, the new electrostatic force would be:
So, the electrostatic force decreases by a factor 9.
Answer:
(a) the final velocity of the cart is 0.857 m/s
(b) the impulse experienced by the package is 21.43 kg.m/s
(c) the fraction of the initial energy lost is 0.71
Explanation:
Given;
mass of the package, m₁ = 10 kg
mass of the cart, m₂ = 25 kg
initial velocity of the package, u₁ = 3 m/s
initial velocity of the cart, u₂ = 0
let the final velocity of the cart = v
(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
10 x 3 + 25 x 0 = v(10 + 25)
30 = 35v
v = 30 / 35
v = 0.857 m/s
(b) the impulse experienced by the package;
The impulse = change in momentum of the package
J = ΔP = m₁v - m₁u₁
J = m₁(v - u₁)
J = 10(0.857 - 3)
J = -21.43 kg.m/s
the magnitude of the impulse experienced by the package = 21.43 kg.m/s
(c)
the initial kinetic energy of the package is calculated as;
the final kinetic energy of the package;
the fraction of the initial energy lost;