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3 years ago

9

An electron with a speed of 0.95c is emitted by a supernova, where cc is the speed of light. What is the magnitude of the moment

um of this electron?

1 answer:

krok68 [10]3 years ago

3 0

Answer:

2.59×10¯²² Kgm/s

Explanation:

Data obtained from the question include:

Velocity of electron = 0.95c

Momentum =?

Next, we shall determine the velocity of the electron. This can be obtained as follow:

Velocity of electron = 0.95c

Velocity of Light (c) = 3×10⁸ m/s

Velocity of electron = 0.95c

Velocity of electron = 0.95 × 3×10⁸

Velocity of electron = 2.85×10⁸ m/s

Finally, we shall determine the mometum of the electron.

Momentum is simply defined as the product of mass and velocity. Mathematically, it is expressed as:

Momentum = mass x Velocity

Thus, with the above formula, we calculate the momentum of the electron as follow:

Mass of electron = 9.1×10¯³¹ Kg

Velocity of electron = 2.85×10⁸ m/s

Momentum of electron =?

Momentum = mass x Velocity

Momentum = 9.1×10¯³¹ × 2.85×10⁸

Momentum = 2.59×10¯²² Kgm/s

Therefore, the momentum of the electron is 2.59×10¯²² Kgm/s

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Answer: 2200J

Explanation:

M = 44kg

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asambeis [7]

Explanation:

It is given that,

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Separation between slits, $d=2.2\times 10^{-5}\ m$

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$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{nc}{fd})$

$\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})$

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Since F = μAu/y

F/A = μu/y where F/A = force per unit area

Since we are given force per unit area, F/A = 1.962 N per unit area = 1.962 N/m²

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substituting the values of the variables into the equation, we have

μ = F/A ÷ u/y

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2 years ago

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Therefore the total heat exchange is given as

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Our values are given as,

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2 years ago