91Ó°ÊÓ

To find out the speed of water exiting the hole, we will use the equation \(v = \sqrt{2gh}\). Here, \(g = 9.81 \mathrm{m/s^2}\) (gravitational acceleration on Earth) and \(h = 1.80 \mathrm{m}\) (the depth of the hole). By plugging the values into the equation, we get: \(v = \sqrt{2 \times 9.81 \times 1.80}\) \(v = \sqrt{35.316}\) \(v \approx 5.94 \mathrm{m/s}\)

As the vat is filled with gasoline, the gravitational acceleration (\(g\)) remains the same, because we are still on Earth. However, Torricelli's theorem assumes an ideal fluid, which means that the fluid's properties, like density, do not affect the speed of the fluid exiting the hole. So, even though gasoline has a different density than water, the speed of gasoline exiting the hole will be the same as in case (a). Therefore, the speed of gasoline exiting the hole is also \(5.94 \mathrm{m/s}\).

In this case, the vat contains water, but we are on the Moon. The gravitational field strength on the Moon is given as \(1.6\, \mathrm{N/kg}\). To find the gravitational acceleration on the Moon, we can use the equation \(g_\text{Moon} = 1.6\, \mathrm{N/kg}\). Now, we can calculate the speed of water exiting the hole on the Moon by using the equation \(v = \sqrt{2gh}\). Plugging the values into the equation, we get: \(v = \sqrt{2 \times 1.6 \times 1.80}\) \(v = \sqrt{5.76}\) \(v \approx 2.4 \mathrm{m/s}\) In conclusion: (a) The speed of water exiting the hole on Earth is approximately \(5.94 \mathrm{m/s}\). (b) Since Torricelli's theorem assumes an ideal fluid, the speed of gasoline exiting the hole will be the same as in case (a), which is approximately \(5.94 \mathrm{m/s}\). (c) The speed of water exiting the hole on the Moon is approximately \(2.4 \mathrm{m/s}\).