Question

power generation

IT.Caleulatehrti of hen particle's ass the particla ravels of rest mass he fullowingfraction O. 0.5, 0.9 o.999 the prollem ) Compute the kineic energy For particle (Mv) for esch fth velecies alectron dleuteran ate thee

Answer 1

Ralatwly femla m mo pautile mass:9 m o050318 1 m Ji-O.1(0.994987) m o o.5 Ji-Co.5) lesclor5) 11547 o.9 | 2. 29415 (0.4 75 93) mo

Tertac o.999 ( C 22.366 27 md 0.0417 timtic Funpy(ke) 111 12 moc KE x G,242イ10“hev C-3x0m1s -31 ofelecton doutron mod 3,346x 10 Rest mes olu hist mass ) (1. o050379 E= 610)o50374-1

Ethod fmoeatx6. .005037- (ons tont () G=(9.1M1G(510)L218 O.5112148 (0.050374 .6050378-1 0.5119198x # 16 MeV 2.575 1547 0,079085 MeV o.9 JIY (KE) = Gx (29415- 1 o511210 8) .29415

Kerfad O.661595 He 42 1 22.3 GG27 7 E) (2.33 G X 0.9.228 Mev for Ruton J KE(mofx C 6,242110 27 3,346x 10 メ ろメ: 1880.109 KF

Ipaye5 01 0050378 kF= (830 109) (005033-) 9.4716 Hev (ii) 0,5 /154 -v E- 1880.10g ((-EASI) 29o.852 Mr 2.29415 P90. (09 (1. 29415-1) 2433。143 H eレ 22. 36627 po.1o9) (2136627) 40170.916 HoU