Lowest frequency of wave
2 answers:
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The momentum of block B after the collision is -50 kg m/s.
Explanation:
We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system before and after the collision must be conserved, so we can write:
where:
is the momentum of block A before the collision
is the momentum of block B before the collision
is the momentum of block A after the collision
is the momentum of block B after the collision
Solving for , we find:
So, the momentum of block B after the collision is -50 kg m/s.
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Answer:
v = 7.5*10⁶ m/s
Explanation:
While accelerating through a potential difference of 160 V, the electron undergoes a change in the electric potential energy, as follows:
ΔUe = q*ΔV = (-e)*ΔV = (-1.6*10⁻¹⁹ C) * 160 V = -2.56*10⁻¹⁷ J (1)
Due to the principle of conservation of energy, in absence of non-conservative forces, this change in potential energy must be equal to the change in kinetic energy, ΔK:
ΔK = Kf -K₀
As the electron accelerates from rest, K₀ =0.
⇒ΔK =Kf = (2)
From (1) and (2):
ΔK = -ΔUe = 2.56*10⁻¹⁷ J =
where me = mass of the electron = 9.1*10⁻³¹ kg.
Solving for vf:
⇒ vf = 7.5*10⁶ m/s
Answer:
Explanation:
dipole moment = qs = q x s
= charge x charge separation
charge = q
separation between charge = s
half separation l = s / 2
dipole has two charges + q and - q separated by distance s .
Potential at distance x along x axis due to + q
Potential at distance x along x axis due to - q
Total potential
v = v₁ + v₂
Potential at distance y along y axis due to + q
Potential at distance y along y axis due to - q
Total potential
v = v₁ + v₂