A small object has charge Q. Charge q is removed from it and placed on a second small object The two objects are placed I m apar
1 answer:
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Answer:
ΔK = -6 10⁴ J
Explanation:
This is a crash problem, let's start by defining a system formed by the two trucks, so that the forces during the crash have been internal and the moment is preserved
initial instant. Before the crash
p₀ = m v₁ + M 0
final instant. Right after the crash
p_f = (m + M) v
p₀ = p_f
mv₁ = (m + M) v
v =
we substitute
v = 3
v = 1.0 m / s
having the initial and final velocities, let's find the kinetic energy
K₀ = ½ m v₁² + 0
K₀ = ½ 20 10³ 3²
K₀ = 9 10⁴ J
K_f = ½ (m + M) v²
K_f = ½ (20 +40) 10³ 1²
K_f = 3 10⁴ J
the change in energy is
ΔK = K_f - K₀
ΔK = (3 - 9) 10⁴
ΔK = -6 10⁴ J
The negative sign indicates that the energy is ranked in another type of energy
Answer:
B) I1 = 1680 kg.m^2 I2 = 1120 kg.m^2
C) V = 0.84m/s T = 29.92s
D) ω2 = 0.315 rad/s
Explanation:
The moment of inertia when they are standing on the edge:
where M is the mass of the merry-go-round.
I1 = 1680 kg.m^2
The moment of inertia when they are standing half way to the center:
I2 = 1120 kg.m^2
The tangencial velocity is given by:
V = ω1*R = 0.84m/s
Period of rotation:
T = 2π / ω1 = 29.92s
Assuming that there is no friction and their parents are not pushing anymore, we can use conservation of the angular momentum to calculate the new angular velocity:
I1*ω1 = I2*ω2 Solving for ω2:
ω2 = I1*ω1 / I2 = 0.315 rad/s
