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blsea [12.9K]
3 years ago
5

Consider two object has mass m1=5.0kg m2=0.5kg and the specific heat c1=c2=4200j/kg oc respectively assume that first object has

temperature 20oc and second object has temperature 90oc After they place these two object touch each other determine the mixture temperature of the substance
Physics
1 answer:
d1i1m1o1n [39]3 years ago
4 0

Answer: T=26.36^{\circ}C

Explanation:

Given

mass of the first object m_1=5\ kg

mass of the second object m_2=0.5\ kg

specific heat of the two is c_1=c_2=4200\ J/kg/^{\circ}C

The temperature of the first T_1=20^{\circ}

The temperature of the second T_2=90^{\circ}

Heat flows from high temperature to low temperature

Suppose T is the common temperature

m_1c_1(T-T_1)=m_2c_2(T_2-T)\\5\times 4200\times (T-20)=0.5\times 4200\times (90-T)\\5T-100=45-0.5T\\5.5T=145\\T=26.36^{\circ}C

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Answer:2.737 kN

Explanation:

Given

mass of log(m)=205 kg

ramp inclination=30^{\circ}

coefficient of kinetic friction between log and ramp is (\mu _k)=0.9

log has an acceleration of 0.8 m/s^2

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