Answer: The molar heat capacity of aluminum is 
Explanation:
As we know that,
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
.................(1)
where,
q = heat absorbed or released
= mass of water = 130.0 g
= mass of aluminiunm = 23.5 g
= final temperature
= 
= temperature of water = 
= temperature of aluminium = 
= specific heat of water= 
= specific heat of aluminium= ?
Now put all the given values in equation (1), we get
![130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]](https://tex.z-dn.net/?f=130.0%5Ctimes%204.184%5Ctimes%20%28299-296%29%3D-%5B23.5%5Ctimes%20c_2%5Ctimes%20%28299-373%29%5D)
Molar mass of Aluminium = 27 g/mol
Thus molar heat capacity =
8/4 = y/y-x
8y - 8x = 4y
y = 2x
y = 2 x 4
y = 8
Hope this helps
Answer:
The answer to your question is: F = 0.4375 N. The force will be 16 times lower than with the first conditions.
Explanation:
Data
F = 7 N
F = ? if the masses is quartered
Formula
Process
Normal conditions F = Km₁m₂/r² = 7
When masses quartered F = K(m₁/4)(m₂/4)/r² = ?
F = K(m₁m₂/16)/r²
F = K(m₁m₂/16r² = 7/16 = 0.4375 N
Answer:
It is an example of velocity
Explanation:
It is an example of velocity Don't ask how I know because I do know it I just don't know how to explain it.
