At that temperature, water is a gas (Steam) unless it is under a huge amount of pressure to keep it liquid, and will have a density that depends on the container that it is in. Saturated steam at that temperature, where it is as dense as possible without turning back into water, will have a density of only 28.604 kg/ m^3. I found this using a saturated steam table. If you use the website I've provided, they will give you the answer in specific volume, which is the inverse of density, so you'll have to flip the fraction upside down and then convert to metric units using the OTHER site I provided.
You go by means of derivation of v = [2*g*h]^(half of) i.E., v = rectangular root of (2gh) it may be achieved by way of balancing knowledge and Kinetic Energies additionally. V = sqrt ( 2 * 9.Eight * 30) = sqrt ( 588) = 24.24 m/s i.E., aprox ....... B.) is 24 m/s
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At that temperature, water is a gas (Steam) unless it is under a huge amount of pressure to keep it liquid, and will have a density that depends on the container that it is in. Saturated steam at that temperature, where it is as dense as possible without turning back into water, will have a density of only 28.604 kg/ m^3. I found this using a saturated steam table. If you use the website I've provided, they will give you the answer in specific volume, which is the inverse of density, so you'll have to flip the fraction upside down and then convert to metric units using the OTHER site I provided.
Best of luck!
You go by means of derivation of v = [2*g*h]^(half of) i.E., v = rectangular root of (2gh) it may be achieved by way of balancing knowledge and Kinetic Energies additionally. V = sqrt ( 2 * 9.Eight * 30) = sqrt ( 588) = 24.24 m/s i.E., aprox ....... B.) is 24 m/s