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Dmitry_Shevchenko [17]

3 years ago

5

Un protón se acelera por una diferencia de potencial de 2.6k V y se dirige a una región donde existe un campo magnético uniforme

de 4.5T. ¿Cuál es el valor?

1 answer:

Bond [772]3 years ago

7 0

Answer:

5.08*10^{-13}N

Explanation:

With this information we can calculate the velocity of the proton by taking into account the kinetic energy of the proton:

$qV=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(1.6*10^{-19}C)(2.6*10^3V)}{1.67*10^{-27}kg}}=705835.3m/s$

The magnitude of the magnetic force will be:

$F=qvB=(1.6*10^{-19}C)(705835.3m/s)(4.5T)=5.08*10^{-13}N$

hope this helps!

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A 70.0 kg70.0 kg ice hockey goalie, originally at rest, has a 0.170 kg0.170 kg hockey puck slapped at him at a velocity of 41.5

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Explanation:

M = mass of ice hockey goalie = 70 kg

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Answer:

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