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3 years ago

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A charged ball is moving horizontally and perpendicular to a magnetic field of 0.8 Tesla. The ball has a mass of 0.007 kg and ha

s a charge of -0.005 C. How fast must the ball be moving in order to cancel out the effect of gravity? Give the velocity as a positive number.

2 answers:

goblinko [34]3 years ago

8 0

Answer:

17.15 m/s

Explanation:

Parameters given:

Magnetic field, B = 0.8 T

Mass of ball, m = 0.007 kg

Charge of ball, q = 0.005 C

The magnetic force acting on the charged ball due to the magnetic field is given as:

F = qvBsinθ

where v = velocity of the ball and θ = angle between the horizontal and the magnetic field = 90°

The force of the ball will be in the opposite direction but of equal magnitude:

$F_b$ = -qvBsin(90) = -qvB

To cancel out the effect of gravity, the magnetic force must be equal to the gravitational force acting on the ball:

F = mg

Therefore:

mg = -qvB

Solving for velocity, v, we have:

$v = \frac{mg}{-qB}$

$v = \frac{0.007 * 9.8}{-(-0.005) * 0.8}$

v = 17.15 m/s

The ball must be moving at a velocity of 17.15 m/s.

mezya [45]3 years ago

7 0

Answer:

8.0

Explanation:

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garri49 [273]

Answer:

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Explanation:

<u>Data</u>

<u>mass m= 100g</u>

<u>Length L= 5cm</u>

<u>we can use:</u>

<u>gm-kL= 0</u>

<u>divide both side by m</u>

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