Physics: A pebble is dropped into a calm pond, causing ripples in the form of concentric circles. The radius of the outer ripple is increasing at a constant rate of 6 inches per second. When the radius is 4 feet, at what rate (in ft2/sec) is the total area A of the disturbed water changing

A pebble is dropped into a calm pond, causing ripples

vianca15

22.01.2022

The rate of area of the disturbed water changing is 12.56 $\frac{ft^{2} }{sec}$

Explanation:

Given:

Radius increasing rate $\frac{dr}{dt} = 6$ $\frac{in}{sec}$

Radius r = 4 ft

Now we convert radius increasing rate into feet per second,

$\frac{dr}{dt} = \frac{6}{12} \frac{ft}{sec}$

Here we have to find total area rate $\frac{dA}{dt}$

$A = \pi r^{2}$

$\frac{dA}{dt} = \frac{d(\pi r^{2} )}{dt}$

$\frac{dA}{dt} =\pi \frac{d( r^{2} )}{dt}$

$\frac{dA}{dt} = 2\pi r\frac{dr }{dt}$

$\frac{dA}{dt} = 2\pi \times 4 \times \frac{6}{12}$

$\frac{dA}{dt} = 12.56$ $\frac{ft^{2} }{sec}$

Therefore, the rate of area of the disturbed water changing is 12.56 $\frac{ft^{2} }{sec}$

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manaspercy

23.11.2022

72 is the premimum of the insurance.

Explanation:

Below is the given values:

The loading = 0.4

Coinsurance rate = 0.2

Number of units = 100

Total number of units = 100 * 0.4 = 40

Remaining units = 60 * 0.2 = 12

Add the 60 and 12 values = 60 + 12 = 72

Thus, 72 is the premimum of the insurance.

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