A proton of mass mp = 1.67 x 10-27kg and charge +e = 1.60 x 10-19...
Homework Answers
energy the proton acquires=1/2mv*v=210000* 1.60 x10-19
so velocity=√2*210000* 1.60 x 10-19/1.67 x10-27=6.3434*106m/sec
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A proton of mass mp = 1.67 x 10-27kg and charge +e = 1.60 x 10-19...
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A proton of mass m = 1.67 x 10-27 kg and charge q = 1.60 x 10-19 C is accelerated from rest through a potential difference of 4000 V. What final velocity does it achieve?
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A proton of mass m = 1.67 x 10-27 kg and charge q = 1.60 x 10-19 C is accelerated from rest through a potential difference of 4000 V. What final velocity does it achieve?
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If a proton m = 1.67 x 10-27 kg and charge q = 1.60 x 10-19 C is released from rest at the + plate in the below electric field, what will be its speed at the-plate? 3 cm E = 5000 N/C
(#966932-48) (Motion of Charge in Field) D2 (Motion Perpendicular to Field) A proton of mass mp-1.67 × 10-27 kg and a charge of gp-1.60 x 10-19 C is moving through vacuum at a constant velocity of 10, 000 m/s directly to the east when it enters a region of uniform electric field that points to the south with a magnitude of El =2.83e+3 N/C . The region of uniform electric field is 5 mm wide in the east-west direction. How...
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