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3 years ago

What is the magnetic force on a proton that is moving at 5.2 x 107 m/s to the

Physics

2 answers:

alisha [4.7K]3 years ago

6 0

Answer:

1.1648×10⁻¹¹ N

Explanation:

Using

F = qvBsinФ..................... Equation 1

Where F = Force on the proton, q = charge, v = velocity, B = magnetic Field, Ф = angle between the magnetic Field and the velocity.

Note: The angle between v and B = 90°

Given: v = 5.2×10⁷ m/s, B = 1.4 T, q = 1.6×10⁻¹⁹ C, Ф = 90°

Substitute into equation 1

F = 1.6×10⁻¹⁹(5.2×10⁷)(1.4)sin90°

F = 11.648×10⁻¹²

F = 1.1648×10⁻¹¹ N.

kogti [31]3 years ago

6 0

Answer:

2.0 x 10^-11

Explanation:

idk if its up or down but look in the comments of this answer, i will put it there when i find out

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$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

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PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

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$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

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$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

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