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natali 33 [55]
4 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?
Physics
1 answer:
krok68 [10]4 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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If you are confused about how I found the acceleration; I plugged in 2.5 for the final y value, 0 for the initial y value, 0 for the initial velocity and 4 for t squared.

To solve, for acceleration it’s a matter of simple algebra. You can subtract the initial y position and the initial velocity from the final y position because they are 0. This leaves you with 2.5 m = 1/2a * t^2, from here I multiplied 2.5 by 2 to get rid of the 1/2. Now I have 5 = a * t^2. T^2 is just 2 squared, so four. Simply divide 5 by 4, and boom, you get 1.25 m/s^2.

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3 years ago
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