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:
Answer:
30×6=n
Explanation:
30 students in her class× 6 = number of students in the entire 4th grade.
Answer:
179 degree
Explanation:
Solid copper will lose heat to warm up and melt ice . Since all the ice has not melted so final temperature of mixture will be zero.
heat gained by ice to reach temperature of zero degree
= mass of ice x specific heat of ice x rise of temp
= 2 x 2093 x 16 = 66976 J
Heat gained to melt 1.1. kg of ice
mass x latent heat of ice
1.1 x 334000
= 367400
total heat
= 367400 + 66976
=434376
Heat lost by copper
= 6.3 x 385 x T
where 385 is specific heat of copper and T is initial temperature
heat lost = heat gained
6.3 x 385 T = 434376
T = 434376 / (6.3 x 385)
= 179 degree .