3 years ago

The minimum number of vectors of unequal magnitude required to

Physics

1 answer:

EleoNora [17]3 years ago

4 0

Answer:

Explanation:

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ra1l [238]

Answer:5000000000hertz

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A particle with charge q = 10-9 C and mass m = 5.0 x 10-9 kg is moving in a magnetic field whose magnitude is 0.003 T. The speed

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Answer:

a=0.212 m/s²

Explanation:

Given that

q= 10⁻⁹ C

m = 5 x 10⁻⁹ kg

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θ= 45°

Lets take acceleration of the mass is a m/s²

The force on the charge due to magnetic filed B

F= q V B sinθ

Also F= m a ( from Newton's law)

By balancing these above two forces

m a= q V B sinθ

$a=\dfrac{qVB\ sin\theta}{m}$

$a=\dfrac{10^{-9}\times 500\times 0.003\times\ sin45^{\circ}}{5\times 10^{-9}}\ m/s^2$

$a=\dfrac{10^{-9}\times 500\times 0.003\times\dfrac{1}{\sqrt2}}{5\times 10^{-9}}\ m/s^2$

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The answer is b !!!! Hope it helps

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Equation is already balanced.

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