Answer:
Temperature increase = 2.1 [C]
Explanation:
We need to identify the initial data of the problem.
v = velocity of the copper sphere = 40 [m/s]
Cp = heat capacity = 387 [J/kg*C]
The most important data given is the fact that when the shock occurs kinetic energy is transformed into thermal energy, therefore it will have to be:
![E_{k}=Q\\ E_{k}= kinetic energy [J]\\Q=thermal energy [J]\\Re-employment values and equalizing equations\\\\\frac{1}{2} *m*v^{2}=m*C_{p}*dT \\The masses are canceled \\\\dT=\frac{v^{2}}{C_{p} *2} \\dT=2.1 [C]](https://tex.z-dn.net/?f=E_%7Bk%7D%3DQ%5C%5C%20E_%7Bk%7D%3D%20kinetic%20energy%20%5BJ%5D%5C%5CQ%3Dthermal%20energy%20%5BJ%5D%5C%5CRe-employment%20values%20and%20equalizing%20equations%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%3Dm%2AC_%7Bp%7D%2AdT%20%20%5C%5CThe%20masses%20are%20canceled%20%5C%5C%5C%5CdT%3D%5Cfrac%7Bv%5E%7B2%7D%7D%7BC_%7Bp%7D%20%2A2%7D%20%5C%5CdT%3D2.1%20%5BC%5D)
Explanation:
M₂ = Fr²/GM₁
M₂ = [(132N)(.243m)²]/[(6.67*10^-11N*m²/kg)(1.175*10^4kg)]
M₂ = (7.79N*m²)/(7.84*10^-7N*m²)
M₂ = 9.94*10^6 kg
Answer:
tbh vector does not have any direction at all the answer is 0
M = 7.0 kg, the mass of the groceries
h = 1.2 m, the elevation of the bag of groceries
The bag of groceries moves a constant velocity over the 2.7-m room.
At constant velocity, there is no applied force, and the kinetic energy remains constant.
At an elevation of 1.2 m, there is an increase in PE (potential energy) given by
V = m*g*h
= (7.0 kg)*(9.8 m/s²)*(1.2 m)
= 82.32 J
The change in PE is equal to the work done.
Answer: 82.3 J
