Answer:

Explanation:
The problem is solved using the law of conservation of energy,
So




Answer : The final temperature 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 ice = 
= specific heat of water = 
= mass of ice = 50 g
= mass of water = 200 g
= final temperature = ?
= initial temperature of ice = 
= initial temperature of water = 
Now put all the given values in the above formula, we get:


Therefore, the final temperature is, 
Answer:
Explanation:
1.2(0) + 3(0.8) + 1.4(0.8/2) / (1.2 + 3 + 1.4) = 0.5285714... ≈ 0.53 m
Answer:
Blue Lighting
Explanation:
In order to make red look black, you must use blue light. The blue would be absorbed and there would be no red light to reflect.