Answer : The heat change of the cold water in Joules is, 
Explanation :
First we have to calculate the mass of cold water.
As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.


Now we have to calculate the heat change of cold water.
Formula used :

where,
Q = heat change of cold water = ?
m = mass of cold water = 45 g
c = specific heat of water = 
= initial temperature of cold water = 
= final temperature = 
Now put all the given value in the above formula, we get:


Therefore, the heat change of cold water is 
Answer:
-5 m/s
Explanation:
The linear velocity of B is equal and opposite the linear velocity of E.
vB = -vE
vB = -ωE rE
10 m/s = -ωE (12 m)
ωE = -0.833 rad/s
The angular velocity of E is the same as the angular velocity of D.
ωE = ωD
ωD = -0.833 rad/s
The linear velocity of Q is the same as the linear velocity of D.
vQ = vD
vQ = ωD rD
vQ = (-0.833 rad/s) (6 m)
vQ = -5 m/s
Answer:

The formula for escape velocity where:
G - Gravitational constant (9.81 etc.)
M - the mass of the object the escape should be made from
r - distance to the centre of that mass
1. Gas
2. Hot
Are the answers.
A because water has more volume when it's warmer