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Rasek [7]
4 years ago
13

For your senior project, you would like to build a cyclotron that will accelerate protons to 10% of the speed of light. The larg

est vacuum chamber you can find is 60 cm in diameter.
Physics
1 answer:
timofeeve [1]4 years ago
6 0

The given question is incomplete. The complete question is as follows.

For your senior project, you would like to build a cyclotron that will accelerate protons to 10% of the speed of light. The largest vacuum chamber you can find is 60 cm in diameter.

What magnetic field strength will you need?

Explanation:

Formula for the strength of magnetic field is as follows.

      B = \frac{mv}{qr}

Here,    m = mass of proton = 1.67 \times 10^{-27} kg

      v = velocity = 10% of 3 \times 10^{8} = 3 \times 10^{7} m/s

      q = charge of proton = 1.6 \times 10^{-19} C

      r = radius = \frac{60}{2} = 30 cm = 0.30 m   (as 1 m = 100 cm)

Therefore, magnetic field will be calculated as follows.

            B = \frac{mv}{qr}

               = \frac{1.67 \times 10^{-27} \times 3 \times 10^{7}}{1.6 \times 10^{-19} C \times 0.30 m}

               = \frac{5.01 \times 10^{-20}}{0.48 \times 10^{-19}}

               = 1.0437 T

Thus, we can conclude that magnetic field strength is 1.0437 T.

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(a) The time taken for the projectile to reach the maximum height is 32.65 s.

(b) The horizontal range of the projectile is 9,049.1 m.

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The time taken for the projectile to reach the maximum height is calculated as follows;

v_f = u- gt\\\\0 = 320 - 9.8t\\\\9.8t = 320\\\\t = \frac{320}{9.8} \\\\t = 32.65 \ s

The horizontal range of the projectile is calculated as follows;

R = \frac{u^2 sin(2\theta)}{g} \\\\R = \frac{320^2 \times sin(2\times 30)}{9.8} \\\\R = 9,049.1 \ m

Learn more about horizontal range here: brainly.com/question/12870645

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