The time needed to empty a vertical cylindrical tank varies directly as the square root of the height of the tank and the square of its radius. By what factor will the emptying time change if the height is doubled and the radius increased by 25%?

t1 = r^2sqrt(h),

t2 = (r+1.25r)^2*sqrt(2h),
t2 = (2.25r)^2*sqrt(2h),
t2 = 5.0625r^2*1.414sqrt(h),
t2 = 7.16r^2*2qrt(h),

t2/t1 = 7.16r^2*sqrt(h) / r^2*2qrt(h) =
7.16.

The emptying time is increased by a factor of 7.16.

Let's denote the original height of the tank as 'h' and the original radius as 'r'. The original emptying time can be represented as t = k * sqrt(h) * r^2, where k is a constant of proportionality.

If the height is doubled, the new height becomes 2h. And if the radius is increased by 25%, the new radius becomes 1.25r.

Plugging these values into the equation, we get the new emptying time:
t' = k * sqrt(2h) * (1.25r)^2

To find the factor by which the emptying time changes, we can divide the new emptying time by the original emptying time:
t' / t = (k * sqrt(2h) * (1.25r)^2) / (k * sqrt(h) * r^2)

Simplifying the equation:
t' / t = (sqrt(2h) * (1.25r)^2) / (sqrt(h) * r^2)
= sqrt(2h) * (1.25r)^2 / sqrt(h) * r^2
= (sqrt(2h) * 1.25^2) / sqrt(h)

Using the fact that (x * y^2) / z = (x/z) * y^2, we can further simplify:
t' / t = (sqrt(2h) / sqrt(h)) * 1.25^2
= sqrt(2) * 1.25^2
= sqrt(2) * 1.5625

Therefore, the emptying time will change by a factor of approximately sqrt(2) * 1.5625 when the height is doubled and the radius is increased by 25%.

To find the factor by which the emptying time changes, we need to understand the relationship between the emptying time and the height and radius of the tank.

According to the problem, the time needed to empty the tank varies directly as the square root of the height and the square of the radius. Let's denote the time needed to empty the tank as T, the original height as h, and the original radius as r.

Therefore, we have the direct proportion equation:

T = k * sqrt(h) * r^2

Where k is the constant of proportionality.

Now, let's consider the changes in the tank's dimensions. The height is doubled, so the new height, which we'll denote as h', is 2h. The radius is increased by 25%, so the new radius, which we'll denote as r', is 1.25r.

Substituting the new values into the equation, we get:

T' = k * sqrt(2h) * (1.25r)^2

Simplifying:

T' = k * sqrt(2h) * 1.25^2 * r^2
T' = k * sqrt(2h) * 1.5625 * r^2
T' = 1.5625 * k * sqrt(2h) * r^2

Now, we can compare the new emptying time, T', to the original emptying time, T, and find the factor by which it changes:

Factor of Change = T' / T
= (1.5625 * k * sqrt(2h) * r^2) / (k * sqrt(h) * r^2)
= 1.5625 * sqrt(2h) / sqrt(h)
= 1.5625 * sqrt(2)

The factor by which the emptying time will change is 1.5625 times the square root of 2.

Therefore, the emptying time will change by a factor of approximately 2.212.