A subatomic particle X spontaneously decays into two particles, A and B, each of rest energy 1.40 × 10^2 MeV. The particles fly
1 answer:
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Answer:
30.7m
Explanation:
You can use the following equation:
ρ = R *
Where ρ is the resistivity, R is the resistance, S is the Cross-sectional Area and l is the length of the wire.
We issolate l in the equation and solve with the data we have:
l = R * = 3.25 Ω *
l = 30.7m
Answer:
Energy stored in the capacitor before the dielectric was inserted=
3.744×10-3 J
Energy stored in the capacitor after the dielectric was inserted=
14.0×10-3 J
Change in energy due to dielectric
10.3×10-3 J or 10mJ
The energy stored in the capacitor increased due to the insertion of the dielectric material.
Explanation:
The energy stored in the capacitor is calculated from the formula
U= 1/2CV^2
Where:
U=energy stored in the capacitor
C=capacitance of the capacitor
V= voltage.
The energy stored in the capacitor before the insertion of the dielectric is shown as U while the energy stored in the capacitor after the insertion of the dielectric is designated U'.
Since U'>U from the solution attached, the change in energy due to insertion of the dielectric ∆U= U'-U
Hence the answer shown. See attached image for detailed solution.
