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The CERN particle accelerator is circular with a circumference of 7.0 km.

(a) What is the acceleration of the protons (m=1.67×10−27kg) that move around the accelerator at 5% of the speed of light? (The speed of light is v=3.00×108m/s.)
(b) What is the force on the protons?

User M Waz
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Final answer:

The proton's acceleration in the CERN particle accelerator is approximately 4.03×1012 m/s2, and the force on the protons due to this acceleration is 6.73×10-15 N.

Step-by-step explanation:

Acceleration and Force on Protons in a Particle Accelerator

Given the CERN particle accelerator's circumference is 7.0 km, we can calculate the radius 'r' using the formula for circumference C = 2πr. Thus, r = C / (2π) = 7000 m / (2π) which is approximately 1114.16 m. The velocity 'v' of the protons at 5% of the speed of light is v = 0.05 × 3.00×108 m/s.

(a) Proton Acceleration: The centripetal acceleration 'a' can be calculated as a = v2 / r. Substituting the given values, a = (0.05 × 3.00×108)2 / 1114.16 m.

Now, calculate the numerical value: a = (0.05 × 3.00×108)2 / 1114.16 m ≈ 4.03×1012 m/s2.

(b) Force on Protons: The force 'F' experienced by the protons is given by Newton's second law, F = ma, where 'm' is the mass of the proton. Using the proton's mass, m = 1.67×10-27 kg, and the calculated acceleration, F = 1.67×10-27 kg × 4.03×1012 m/s2. Calculating this product gives F = 6.73×10-15 N.

User Shakilur
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