Answer:
The time after which the two stones meet is tₓ = 4 s
Explanation:
Given data,
The height of the building, h = 200 m
The velocity of the stone thrown from foot of the building, U = 50 m/s
Using the II equation of motion
S = ut + ½ gt²
Let tₓ be the time where the two stones meet and x be the distance covered from the top of the building
The equation for the stone dropped from top of the building becomes
x = 0 + ½ gtₓ²
The equation for the stone thrown from the base becomes
S - x = U tₓ - ½ gtₓ² (∵ the motion of the stone is in opposite direction)
Adding these two equations,
x + (S - x) = U tₓ
S = U tₓ
200 = 50 tₓ
∴ tₓ = 4 s
Hence, the time after which the two stones meet is tₓ = 4 s
Answer:

Explanation:
It is given that,
Diameter of cylinder, d = 6.6 cm
Radius of cylinder, r = 3.3 cm = 0.033 m
Acceleration of the string, 
Displacement, d = 1.3 m
The angular acceleration is given by :



The angular displacement is given by :



Using the third equation of rotational kinematics as :

Here, 



Since, 1 rad/s = 9.54 rpm
So,

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.
You first find the mass and the volume of that object. Then you divide mass ÷ volume
Answer: (B) There is complete destructive interference between the incoming and reflected waves
Explanation:
For example, if you pluck a guitar the waves will travel back and forth. They consist of nodes and anti-nodes. It is created, when the wave traveling to one side and bounces of the other end and comes back. As it travels to the other side, it is reflected thus, comes back. So standing waves occurs when there is interference.
When the wave is produced, the points where the string is not moving are called nodes and where they are moving are called anti-nodes. The positions where nodes are produced, destructive interference occurs and where anti-nodes are produced, constructive interference occurs