What are three physical of aluminum foil
1 answer:
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Answer:
The final temperature of both objects is 400 K
Explanation:
The quantity of heat transferred per unit mass is given by;
Q = cΔT
where;
c is the specific heat capacity
ΔT is the change in temperature
The heat transferred by the object A per unit mass is given by;
Q(A) = caΔT
where;
ca is the specific heat capacity of object A
The heat transferred by the object B per unit mass is given by;
Q(B) = cbΔT
where;
cb is the specific heat capacity of object B
The heat lost by object B is equal to heat gained by object A
Q(A) = -Q(B)
But heat capacity of object B is twice that of object A
The final temperature of the two objects is given by
But heat capacity of object B is twice that of object A
Therefore, the final temperature of both objects is 400 K.
Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant
The work done in stretching the rubber band from x = 0 to x = L is
![W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2} [x^{2} ]_{0}^{L} = \frac{aL^{2}}{2}](https://tex.z-dn.net/?f=W%3D%5Cint_%7B0%7D%5E%7BL%7D%20Fdx%20%3D%20%5Cint_%7B0%7D%5E%7BL%7Dax%20%5C%2C%20dx%20%3D%20%5Cfrac%7Ba%7D%7B2%7D%20%20%5Bx%5E%7B2%7D%20%5D_%7B0%7D%5E%7BL%7D%20%3D%20%20%5Cfrac%7BaL%5E%7B2%7D%7D%7B2%7D%20)
Answer:
F = ax
where
x = distance by which the rubber band is stretched
a = constant
The work done in stretching the rubber band from x = 0 to x = L is
Answer: