<span>An isotope is a form of a
chemical element whose atomic nucleus contains a specific number of neutrons in
addition to the number of protons that distinctively defines the element. The
nuclei of most atoms have neutrons as well as protons.</span>
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Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
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Maybe you can split up the questions. I will try to answer your first question.
1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)
2. Momentum: p = m₁v₁ + m₂v₂
m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B
Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0
Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0
3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.
4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7
5.You figure out.
