Water is poured into a conical container at the rate of 10 cm3/sec. the cone points directly down, and it has a height of 20 cm and a base radius of 15 cm. how fast is the water level rising when the water is 2 cm deep (at its deepest point)?

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

Of course, .

Now, since the volume of a cone is and the ratio or , the volume becomes:

We proceed to its differentiation:

Then, we compute

Finally, at h=2:

Best regards.

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !