I think it occurs whenever waves come together so that they are in phase with each other.
Answer: 6.47m/s
Explanation:
The tangential speed can be defined in terms of linear speed. The linear speed is the distance traveled with respect to time taken. The tangential speed is basically, the linear speed across a circular path.
The time taken for 1 revolution is, 1/3.33 = 0.30s
velocity of the wheel = d/t
Since d is not given, we find d by using formula for the circumference of a circle. 2πr. Thus, V = 2πr/t
V = 2π * 0.309 / 0.3
V = 1.94/0.3
V = 6.47m/s
The tangential speed of the tack is 6.47m/s
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, 