Answer:
W = -0.480 J
Explanation:
given,
q₁ = 4 μC
q₂ = -4.10 μC
b = 0.381
k = 8.99 × 10⁹ Nm²/C²
W = -0.480 J
Work done by the electric force W = -0.480 J
Ignoring the electron spin what is the largest possible energy difference, if the magnetic field is 2.02 tesla?
1 answer:
Complete Question:
When specially prepared Hydrogen atoms with their electrons in the 6f state are placed into a strong uniform magnetic field, the degenerate energy levels split into several levels. This is the so called normal Zeeman effect.
Ignoring the electron spin what is the largest possible energy difference, if the magnetic field is 2.02 Tesla?
Answer:
ΔE = 1.224 * 10⁻²² J
Explanation:
In the 6f state, the orbital quantum number, L = 3
The magnetic quantum number,
The change in energy due to Zeeman effect is given by:
Magnetic field B = 2.02 T
Bohr magnetron,
ΔE = 1.224 * 10⁻²² J
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that gives you 20m after 1st second, 40 m after 2nd second, 60 m after 3rd second and so on.
For counting y you have to use the equation for the distanced covered by the object moving with constantly accelerating velocity (symbols refering to vertical movement):
that gives you 5m after 1st second, 20m afters 2nd second, 45m after 3rd second and so on.
Add minus signs before y positions to receive graph presenting the movement of the ball.
So the points are: P1=[20,-5], P2=[40,-20], P3=[60,-45] and so on... Pn=[x,y].
that gives you 20m after 1st second, 40 m after 2nd second, 60 m after 3rd second and so on.
For counting y you have to use the equation for the distanced covered by the object moving with constantly accelerating velocity (symbols refering to vertical movement):
that gives you 5m after 1st second, 20m afters 2nd second, 45m after 3rd second and so on.
Add minus signs before y positions to receive graph presenting the movement of the ball.
So the points are: P1=[20,-5], P2=[40,-20], P3=[60,-45] and so on... Pn=[x,y].
Answer:
p_{f} = 6 m / s
Explanation:
We can solve this exercise using conservation of momentum. For this we define a system formed by the two balls, so that the forces during the collision have been intense and the moment is preserved
Initial instant. Before the crash
p₀ = m v +0
Final moment. Right after the crash
= (m + m) v_{f}
how the moment is preserved
p₀ = p_{f}
m v = 2 m v_{f}
v_{f} = v / 2
we calculate
v_{f} = 12/2
p_{f} = 6 m / s