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A stationary proton is accelerated through a potential difference of 83.5 KV . what speed did...
(c) A proton is accelerated from rest through a potential difference of 10 kV. A proton has chargeqproton = +1.6 10-19 C and mass mproton = +1.67 10-27 C i. Calculate the change in potential energy of the proton, in your answer you must explicitly state whether it is a gain or loss in potential energy. ii. Calculate the final velocity of the proton.
(a) Calculate the speed of a proton that is accelerated from rest through a potential difference of 111 V km/s (b) Calculate the speed of an electron that is accelerated through the same potential difference. Mm/s
A proton is accelerated through a 3.11 kV potential difference and directed between parallel plates separated 12.3 mm as shown below. The EMF of the battery is 10.0 V. What is the magnitude and direction of the uniform magnetic field between the plates that allows the proton to travel undeflected?
A proton (m= 1.67e-27 kg) is accelerated from rest through a potential difference of 11.5 kV before entering a velocity selector. If the B- field of the velocity selector is perpendicular to the velocity and the electric field (E) has a magnitude of 3.5e6 N/C, what is the required magnitude of the magnetic field (B) if the proton is undeflected?
A)A proton is accelerated from rest through a potential difference of 25707 V. What is the kinetic energy of this proton after this acceleration? The mass of a proton is 1.673 × 10−27 kg and the elemental charge is 1.602 × 10−19 C. Answer in units of J. B) What is the speed of the proton after this acceleration? Answer in units of m/s.
(a) Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 158 V. m/s (b) Calculate the speed of an electron that is accelerated through the same potential difference. m/s
A proton is accelerated from rest through a potential difference of 500 Megavolt. a) What is its speed? b) Classical mechanics indicates that quadrupling the potential difference would double the speed. Were the classical analysis valid, what speed would result from a 2000 Megavolt potential difference? c) What speed actually results?
A proton is accelerated by a electric potential of 50 kV. This proton is heading towards a constant magnetic field of 1.2 T perpendicular to its velocity. • Compute the speed of the proton due to the electric potential. • Once the proton is inside the magnetic field region, what is the radius of curvature of its trajectory. • What is the period of revolution?
If an electron is accelerated from rest through a potential difference of 12.0 kV, what is its resulting speed? (e = 1.60 × 10-19 C, k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2, mel = 9.11 x 10-31 kg). (Give your answer to the nearest km/s).
A proton is accelerated by a electric potential of 50 kV. This proton is heading towards a constant magnetic field of 1.2 T perpendicular to its velocity.(12 points) • Compute the speed of the proton due to the electric potential. . Once the proton is inside the magnetic field region, what is the radius of curvature of its trajectory • What is the period of revolution?