Answer:
K = 9.53 MeV
Explanation:
The kinetic energy that the alpha particle has emitted, is the energy in excess after removing the resting energy of the atoms and the helium nucleus that forms the alpha particle
Since energy and masses are related and cannot be
m₀ c² = c² + m_He c²+ K
K = c² (m₀ - m_{f} - m_He)
the mass of the Helium atom is 4 u
K = (3 10⁸)² (211,988868 -207.976652 - 4,002) 1,661 10⁻²⁷
K = 14,949 10⁻¹¹ (0.0102)
K = 1,527 10⁻¹² J
let's reduce 1 J = 6,242 10¹² MeV
K = 9.53 MeV
Answer:
Inductance as calculated is 13.12 mH
Solution:
As per the question:
Length of the coil, l = 12 cm = 0.12 m
Diameter, d = 1.7 cm = 0.017 m
No. of turns, N = 235
Now,
Area of cross-section of the wire, A =
We know that the inductance of the coil is given by the formula:
The percent difference between two numbers and
is given by
The absolute value is there because we only care about the absolute percent difference, and not taking into account whether we go from to
or vice versa. If we remove them, we have two possible interpretations of percent difference.
For example, the (absolute) percent difference between 3 and 6 is
In other words, we add 100% of 3 to 3 to end up with 6. This is the same as the percent difference going from 3 to 6. On the other hand, the percent difference going from 6 to 3 is
which is to say, we take away 50% of 6 away from 6 to end up with 3.
"Make comparisons to object measurements" tells us that the differences should be computed relative to "measurements for object". In other words, take from the left column and
from the right column.