Question

A bullet having a mass of 50 grm moves with a velocity of 1000 mtr/sec find the energy in ergs and in kgm.metres. find answer

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Answer 1

$\huge \boxed{answer}$

$\sf{m = 50 \: g \: \: \: v = 1000m {s}^{ - 1} = 100000cm {s}^{ - 1} } \\ \\ \sf{ke = \frac{1}{2}m {v}^{2} } \\ \\ \: \: \: \: \: = \frac{1}{2} \times 50 \times 100000 \\ \\ \: \: \: \: = \bf{25 \times {10}^{5} \: ergs} \\ \\ \star \sf \: 1 \: erg \: = {10}^{ - 7} joule(kg \: {m}^{2} \: {s}^{ - 2}) \\ \\ \therefore \sf \: 25 \times {10}^{5} erg = 0.25 \: kg \: {m}^{2} \: {s}^{ - 2}$

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Answer 2

$\Large\underline{\underline{\sf \blue{Given}:}}$

• Mass of the bullet =50 grams.
• velocity of the bullet =1000 mtr/sec

$\Large\underline{\underline{\sf \blue{To\:Find}:}}$

• Find the energy in ergs and in Kg.metres.

$Mass = 50\:gm,\:\:\:Velocity = 1000 \: meters/sec\\\\\: or \:\:{10}^{5}cm/sec \:\:\:\:\: \: (\because \: 1 \: m = 100 \: cm)\\\\\leadsto \:Now,\:K.E = \frac{1}{2}mv^{2} = \frac{1}{2} \times 50 \times (10^{5})^{2} = 25 \times {10}^{10} ergs.\\\\ \implies\:Here,\:K.E = {\frac 25\times10^{10}}\:{980} \: gm.\: cms.\:\: \: \: \: \: \: ( \because1gm.cms = 980 \: ergs) \\\\\implies \frac{25 \times 10^{10}}{980 \times 1000 \times 100}\:kgm \times \:metres = 2.55 \times 10^{3} \: kgm \: metre \\\\ = \boxed{ 2.55 \times 10^{3}kgm.metres} \:$