Answer:
After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec
Explanation:
Area of a circle = πr²
where;
r is the circle radius
Differentiate the area with respect to time.
dr/dt = 4 ft/sec
after 12 seconds, the radius becomes = 
To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt
dA/dt = 1206.528 ft²/sec
Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec
Velocity=frequency(wavelength)
24m/s=f(2m)
24/2=f(2)/2
12Hz=f
If l and m both are doubled then the period becomes √2*T
what is a simple pendulum?
It is the one which can be considered to be a point mass suspended from a string or rod of negligible mass.
A pendulum is a weight suspended from a pivot so that it can swing freely.
Here,
A certain frictionless simple pendulum having a length l and mass m
mass of pendulum = m
length of the pendulum = l
The period of simple pendulum is:
Where k is the constant.
Now the length and mass are doubled,
m' = 2m
l' = 2l
Hence,
If l and m both are doubled then the period becomes √2*T
Learn more about Simple Harmonic Motion here:
<u>brainly.com/question/17315536</u>
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Distance and time, distance because that's how far and time because that's how long
