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A 5.00-g lead BB moving at 44.0 m/s penetrates a wood block and comes to rest inside the block. If half of its kinetic energy is absorbed by the BB, what is the change in the temperature of the BB? The specific heat of lead is 128 J/kg ∙ K

Respuesta :

Khoso123

Answer:

ΔT=3.781 Kelvin

Explanation:

Given data

m=5.00g 0r 0.005kg

v=44.0 m/s

c (specific heat)=128 j/kg

Kinetic Energy=[tex]\frac{1}{2}mv^{2}[/tex]

KE=[tex]\frac{1}{2}(0.005)(44)^{2}[/tex]

KE=4.84 joules

1/2 Kinetic energy goes to heat=1/2×4.84

                                                    =2.42 J

Q=mcΔT

ΔT=[tex]\frac{Q}{mc}[/tex]

ΔT=[tex]\frac{2.42}{0.005*128}[/tex]

ΔT=3.781 K

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