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arsen [322]
3 years ago
15

A muon is traveling at 0.988

Physics
1 answer:
AlexFokin [52]3 years ago
6 0

As per Einstein's relation of relativity

m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}

here we know that

m_0 = 207 m_e = 207 \times 9.11 \times 10^{-31} kg

m_0 = 1.88 \times 10^{-28} kg

now here we know that

v = 0.988 c

now from above equation mass of the muon is given as

m = \frac{1.88 \times 10^{-28}}{\sqrt{1 - 0.988^2}}

m = 1.22 \times 10^{-27} kg

now for the momentum of muon we can use

P = mv

P = 1.22 \times 10^{-27} \times 0.988(3 \times 10^8)

P = 3.62 \times 10^{-19} kg m/s

so above is the momentum of muon

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2) Therefore, take into account that the main points to draw a parabola are:

i) vertex

ii) concavity

iii) y-intercepts

iv) x - intercepts

Also, for all graphs you need the domain and the range.

3) Find the y-intercept (t = 0)

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6) Vertex

It is the local minimum of the equation. You can find it by the first derivative

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7) The domain is given t ∈ [0,5.00]

8) You can also build a table with several points in the domain

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t = 1; Vₓ = 3.00 + 0.100 (1)² = 3.10

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