Nah gaya seperti titik fashion seperti apa yang Anda mana dan hal-hal seperti itu. (tell me if you cant understand)
The force of gravity between two objects is:
F = G*m1*m2/r^2
So, it is dependent of the two masses and the distance between their centers of mass.
<u>Answer:</u> The final temperature of the solution is 
<u>Explanation:</u>
The amount of heat released by coffee will be absorbed by aluminium spoon.
Thus, 
To calculate the amount of heat released or absorbed, we use the equation:

Also,
..........(1)
where,
q = heat absorbed or released
= mass of aluminium = 39 g
= mass of coffee = 166 g
= final temperature = ?
= temperature of aluminium = 
= temperature of coffee = 
= specific heat of aluminium = 
= specific heat of coffee= 
Putting all the values in equation 1, we get:
![39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]](https://tex.z-dn.net/?f=39%5Ctimes%200.904%5Ctimes%20%28T_%7Bfinal%7D-24%29%3D-%5B166%5Ctimes%204.1801%5Ctimes%20%28T_%7Bfinal%7D-83%29%5D)

Hence, the final temperature of the solution is 
Answer:
The velocity of water at the bottom, 
Given:
Height of water in the tank, h = 12.8 m
Gauge pressure of water, 
Solution:
Now,
Atmospheric pressue, 
At the top, the absolute pressure, 
Now, the pressure at the bottom will be equal to the atmopheric pressure, 
The velocity at the top,
, l;et the bottom velocity, be
.
Now, by Bernoulli's eqn:

where

Density of sea water, 



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