Answer : The temperature when the water and pan reach thermal equilibrium short time later is, 
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.


where,
= specific heat of aluminium = 
= specific heat of water = 
= mass of aluminum = 0.500 kg = 500 g
= mass of water = 0.250 kg = 250 g
= final temperature of mixture = ?
= initial temperature of aluminum = 
= initial temperature of water = 
Now put all the given values in the above formula, we get:


Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, 
Answer:
Speed is solved with time and distance but has no direction
Average velocity is solved with Δx/Δt and has a direction
Answer: 1175 J
Explanation:
Hooke's Law states that "the strain in a solid is proportional to the applied stress within the elastic limit of that solid."
Given
Spring constant, k = 102 N/m
Extension of the hose, x = 4.8 m
from the question, x(f) = 0 and x(i) = maximum elongation = 4.8 m
Work done =
W = 1/2 k [x(i)² - x(f)²]
Since x(f) = 0, then
W = 1/2 k x(i)²
W = 1/2 * 102 * 4.8²
W = 1/2 * 102 * 23.04
W = 1/2 * 2350.08
W = 1175.04
W = 1175 J
Therefore, the hose does a work of exactly 1175 J on the balloon
The centripetal acceleration is given by

where v is the tangential speed and r the radius of the circular orbit.
For the car in this problem,

and r=40 m, so we can re-arrange the previous equation to find the velocity of the car: