Explanation:
It is given that,
The volume of a right circular cylindrical, 
We know that the volume of the cylinder is given by :

............(1)
The upper area is given by :



For maximum area, differentiate above equation wrt r such that, we get :



r = 1.83 m
Dividing equation (1) with r such that,



Hence, this is the required solution.
Answer:
Explanation:
Let the amplitude of individual wave be I and resultant amplitude be 1.703 I . Let the phase difference be Ф in terms of degree
From the formula of resultant vector
(1.703I)² = I² + I² + 2 I² cosФ
2.9 I² = 2I² + 2 I² cosФ
.9I² = 2 I² cosФ
cosФ = .9 / 2
= .45
Ф = 63.25 .
Answer: No, water in the ocean wouldn't have tides wouldn't be as strong anymore.
Explanation:
Answer:
The object will travel at the speed of 16 m/s.
Explanation:
Given
To determine
How fast is the object traveling?
<u>Important Tip:</u>
The product of the mass and velocity of an object — momentum.
Using the formula

where
Thus, in order to determine the speed of the object, all we need to do is to substitute p = 64 and m = 4 in the formula


switch the equation

divide both sides by 4

simplify
m/s
Therefore, the object will travel at the speed of 16 m/s.
Answer:
When a tree is vigorously shaken, the branches of the tree come in motion but the leaves tend to continue in their state of rest due to inertia of rest. As a result of this, leaves get separated from the branches of the tree and hence fall down.