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mylen [45]
3 years ago
6

A subatomic particle X spontaneously decays into two particles, A and B, each of rest energy 1.40 × 10^2 MeV. The particles fly

off in opposite directions, each with speed 0.827c relative to an inertial reference frame S. What is the total energy of particle A?
Physics
1 answer:
adoni [48]3 years ago
6 0

Answer:

E = 389 MeV

Explanation:

The total energy of particle A, will be equal to the sum of rest mass energy and relative energy of particle A. Therefore,

Total Energy of A = E = Rest Mass Energy + Relative Energy

Using Einstein's Equation: E = mc²

E = m₀c² + mc²

From Einstein's Special Theory of Relativity, we know that:

m = m₀/[√(1-v²/c²)]

Therefore,

E = m₀c² + m₀c²/[√(1-v²/c²)]

E = m₀c²[1 + 1/√(1-v²/c²)]

where,

m₀c² = rest mass energy = 140 MeV

v = relative speed = 0.827 c

Therefore,

E = (140 MeV)[1 + 1/√(1 - (0.827c)²/c²)]

E = (140 MeV)(2.78)

<u>E = 389 MeV</u>

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An object is formed by attaching a uniform, thin rod with a mass of mr = 6.85 kg and length L = 5.76 m to a uniform sphere with
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Answer:

Part a)

I = 1879.7 kg m^2

Part b)

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Part c)

I = 153.8 kg m^2

Part 4)

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I = 345.6 kg m^2

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Part a)

Moment of inertia of the system about left end of the rod is given as

I = \frac{m_r L^2}{3} + (\frac{2}{5} m_s R^2 + m_s(R + L)^2)

So we have

I = \frac{m_r(4R)^2}{3} + (\frac{2}{5}(5m_r) R^2 + (5m_r)(R + 4R)^2)

I = \frac{16}{3}m_r R^2 + (2m_r R^2 + 125 m_rR^2)

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Part b)

If force is applied to the mid point of the rod

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\tau = F\frac{L}{2}

\tau = 460(2R)

\tau = 460 \times 2 \times 1.44

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Part c)

position of center of mass of rod and sphere is given from the center of the sphere as

x = \frac{m_r}{m_r + m_s}(\frac{L}{2} + R)

x = \frac{m_r}{6 m_r}(3R) = \frac{R}{2}

so moment of inertia about this position is given as

I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + \frac{R}{2})^2 + (\frac{2}{5} m_s R^2 + m_s(\frac{R}{2})^2)

so we have

I = \frac{m_r (16R^2)}{12} + m_r(\frac{5R}{2})^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(\frac{R^2}{4})

I = m_r R^2(\frac{16}{12} + \frac{25}{4} + 2 + \frac{5}{4})

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moment of inertia about right edge of the sphere is given as

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I = \frac{m_r (16R^2)}{12} + m_r(4R)^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(R^2)

I = m_r R^2(\frac{16}{12} + 16 + 2 + 5)

I = 6.85(1.44)^2\times 24.33

I = 345.6 kg m^2

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