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Helga [31]
3 years ago
15

A proton in a cyclotron is moving with a speed of 2.97×107 m/s in a circle of radius 0.568 m. 1.67 × 10−27 kg is the mass of the

proton, and 1.60218 × 10−19 C is its fundamental charge. What is the magnitude of the force exerted on the proton by the magnetic field of the cyclotron? Answer in units of N.
Physics
1 answer:
vivado [14]3 years ago
3 0

Answer:

B = 0.546 T,  F = 2.59 10⁻¹² N

Explanation:

The magnetic force is

            F = q v x B

We can calculate the magnitude of the force and find the direction by the right hand rule

          F = q v B sin θ

Let's use Newton's second law

         F = m a

Acceleration is centripetal

         a = v² / r

We substitute

       q v B sin θ = m v² / r

The angle between the field and the radius of the circle is 90º so sin 90 = 1

        q B = m v / r

        B = m v / q r

Let's calculate ’

       B = 1.67 10⁻²⁷ 2.97 10⁷ / (1.60 10⁻¹⁹ 0.568)

        B = 0.546 T

The foce is

         F = q v B

         F = 1.60 10⁻¹⁹ 2.97 10⁷ 0.546

         F = 2.59 10⁻¹² N

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A phonograph record 0.15 m in its radius rotates 18 times per 90 seconds what is the frequency?
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Answer:

The frequency of the phonograph record is 0.2 Hz

Explanation:

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The given parameters of the phonograph record are;

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2 years ago
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Answer:

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Explanation:

We can start by the time it takes for the driver to come from 22.8m/s to full rest:

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Using the following equation motion we can findout the angle it makes during the deceleration:

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where \omega = 0 m/s is the final angular velocity of the car when it stops, \omega_0 = 114rad/s is the initial angular velocity of the car \alpha = 14.75 rad/s2 is the deceleration of the can, and \Delta \theta is the angular distance traveled, which we care looking for:

-114^2 = 2*(-14.75)*\Delta \theta

\Delta \theta = 440rad or 440/2π = 70 revelutions

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