How do I find the velocity that an electron must acquire

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The discussion centers on calculating the velocity an electron must achieve to match the rest mass of a proton. The relevant formula used is m = m_0 / (1 - v^2/c^2)^(1/2), where m is the relativistic mass, m_0 is the rest mass, and c is the speed of light. The masses of the electron (9.11 x 10^-31 kg) and proton (1.67 x 10^-27 kg) are provided, leading to the conclusion that the velocity can be approximated using v = (√(1836² - 1) / 1836) c. This result indicates that the required velocity is nearly equal to the speed of light, consistent with principles of Special Relativity (SR).

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How do I find the velocity that an electron must acquire, if its mass is to be equal to the rest mass of a proton?

I know that the masses of electron and proton are 9.11 X 10^-31 kg and 1.67 X 10^-27 kg, respectively.

Am I going to use the formula m = m_0 / (1 - v^2/c^2)^(1/2) ?

How do I continue from here?
 
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irony of truth said:
How do I find the velocity that an electron must acquire, if its mass is to be equal to the rest mass of a proton?

I know that the masses of electron and proton are 9.11 X 10^-31 kg and 1.67 X 10^-27 kg, respectively.

Am I going to use the formula m = m_0 / (1 - v^2/c^2)^(1/2) ?

How do I continue from here?

Let's approximate this number,okay??
\frac{9.11\cdot 10^{-31}}{1.67\cdot 10^{-27}}\sim \frac{1}{1836}

I'll give u the final answer and let u work out your way through to it.
v=\frac{\sqrt{1836^{2}-1}}{1836} c

Good luck!
 
Hmm, I just got your reply.. seems like it is almost the same as the speed of light... amazing!
 
irony of truth said:
Hmm, I just got your reply.. seems like it is almost the same as the speed of light... amazing!

Judging it in the context of SR,it's not surprising,let alone amazing.So,by "I just got your reply" does it mean you solved the problem and confronted the result with mine ??Hopefully so,it would mean that the PF community did its job and got u happy and satisfied... :approve: :approve:
 
So,by "I just got your reply" does it mean you solved the problem and confronted the result with mine ??

Hi! What I mean is that when I saw your hints, I quickly solve the problem - knowing that my first idea is correct... but I did not confront your answer... I just saw that I got the correct solution.. same as yours.

But I thank you for helping me.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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