heat released Q = 749 joules
heat of fusion of silver L = 109 J/g
Here phase of silver is changing from liquid to solid
so temperature will remain same
all heat will be released due to its phase change
and in this case we use Q=mL
where m is the mass of silver in gram
Q= mL
749 = m * 109
m = 749/109
m = 6.87 gram
Answer:
l think C am not pretty show
Answer with Explanation:
We are given that
Mass of spring,m=3 kg
Distance moved by object,d=0.6 m
Spring constant,k=210N/m
Height,h=1.5 m
a.Work done to compress the spring initially=![\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dkx%5E2%3D%5Cfrac%7B1%7D%7B2%7D%28210%29%280.6%29%5E2%3D37.8J)
b.
By conservation law of energy
Initial energy of spring=Kinetic energy of object
![37.8=\frac{1}{2}(3)v^2](https://tex.z-dn.net/?f=37.8%3D%5Cfrac%7B1%7D%7B2%7D%283%29v%5E2)
![v^2=\frac{37.8\times 2}{3}](https://tex.z-dn.net/?f=v%5E2%3D%5Cfrac%7B37.8%5Ctimes%202%7D%7B3%7D)
![v=\sqrt{\frac{37.8\times 2}{3}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7B37.8%5Ctimes%202%7D%7B3%7D%7D)
v=5.02 m/s
c.Work done by friction on the incline,![w_{friction}=P.E-spring \;energy](https://tex.z-dn.net/?f=w_%7Bfriction%7D%3DP.E-spring%20%5C%3Benergy)
![W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J](https://tex.z-dn.net/?f=W_%7Bfriction%7D%3D3%5Ctimes%209.8%5Ctimes%201.5-37.8%3D6.3%20J)
They will rise to the 2nd layer of the atmosphere where the temperature decreases by a lot and then they will blow up