A light horizontal spring has a spring constant of 167 N/m. A 2.77 kg block is pressed against one end of the spring, compressin
1 answer:
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M = 20.0kg, the mass of the block.
c = 1700 J/(kg-°C), the specific heat
ΔT = 25 - 15 = 10 °C = 10 K, the change in temperature.
The change in thermal energy is
Answer: 340 kJ (or 340,000 J)
c = 1700 J/(kg-°C), the specific heat
ΔT = 25 - 15 = 10 °C = 10 K, the change in temperature.
The change in thermal energy is
Answer: 340 kJ (or 340,000 J)
Answer:
ΔS=2*m*Cp*ln((T1+T2)/(2*(T1*T2)^1/2))
Explanation:
The concepts and formulas that I will use to solve this exercise are the integration and the change in the entropy of the universe. To calculate the final temperature of the water the expression for the equilibrium temperature will be used. Similarly, to find the change in entropy from cold to hot water, the equation of the change of entropy will be used. In the attached image is detailed the step by step of the resolution.
Answer:
K = 80.75 MeV
Explanation:
To calculate the kinetic energy of the antiproton we need to use conservation of energy:
<em>where : is the photon energy,
: are the rest energies of the proton and the antiproton, respectively, equals to m₀c²,
: are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.</em>
Therefore the kinetic energy of the antiproton is:
<u>The proton mass is equal to the antiproton mass, so</u>:
Hence, the kinetic energy of the antiproton is 80.75 MeV.
I hope it helps you!