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A lead bulletc just melts when stopped by an obstacle .bulle absorb 75%heat.find minimum speed of bullet.initial temperature is 27degree Celsius, specific heat capacity =0.126j/g/degreek Celsius and latent heat of fusion=25.2 /get.Here why do we multiply 10 power _3 to 1/2mv^2 while finding kinetic energy . the specific heat capacity is given in gram ,but in 1/2mv^2,we don’t need it.we want only mass

A lead bulletc just melts when stopped by an obstacle .bulle absorb 75%heat.find minimum speed of bullet.initial temperature is 27degree Celsius, specific heat capacity =0.126j/g/degreek Celsius and latent heat of fusion=25.2 /get.Here why do we multiply 10 power _3 to 1/2mv^2 while finding kinetic energy .the specific heat capacity is given in gram ,but in 1/2mv^2,we don’t need it.we want only mass

Grade:11

1 Answers

Arun
25763 Points
2 years ago
Explanation: K.E. of bullet = (1/2) mV2. After striking KE is converted to heat. Hence Q = (1/2) mV2. Given is that 25% heat is absorbed by obstacle, hence Q' = 0.75 × (1/2) mV2      (1) This Q' will melt the bullet. ΔT = 327 – 27 = 300°C. If C is specific heat & L is latent heat then Q' = (mCΔT) + (m × L)  ∴ form (1) [{0.75} / 2] mV2 = mCΔT + mL ∴ 0.375 V2 = C ∙ ΔT + L V2  = [{C ∙ ΔT + L} / {0.375}] = [{(0.03 × 300) + 6} / {0.375}] = 40kcal/kg = 40 × 4.2 × 103 J/kg ∴ V2 = 168000 ∴ V = 409.90 m/s

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