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What is the maximum mass the Flyboard could support if [tex]Q = 100 \, \text{L/s}[/tex] and [tex]D_{\text{hose}} = 0.1 \, \text{m}[/tex], and the exit nozzles are directed downwards from the horizontal plane at 45 degrees?

A. 100 kg
B. 155 kg
C. 195 kg
D. 245 kg
E. 300 kg

Answer :

Final answer:

The maximum mass the Flyboard could support is determined by the force generated by the water propulsion system. It can be calculated using the rate of water flow, the change in momentum of the water, and the angle at which the exit nozzles are directed downwards. The maximum mass supported by the Flyboard is influenced by the force generated by the water propulsion system.

Explanation:

The maximum mass the Flyboard can support can be calculated using the principles of fluid dynamics and Newton's laws of motion.

First, we need to calculate the force generated by the water propulsion system. The force (F) can be calculated using the equation:

F = Q * Δp

Where Q is the rate of water flow and Δp is the change in momentum of the water.

Next, we need to calculate the change in momentum of the water. The change in momentum (Δp) can be calculated using the equation:

Δp = m * Δv

Where m is the mass of water expelled per second and Δv is the change in velocity of the water.

Since the exit nozzles are directed downwards at 45 degrees, the change in velocity (Δv) can be calculated using the equation:

Δv = 2 * v * sin(45)

Where v is the velocity of the water.

Finally, we can calculate the maximum mass (m_max) the Flyboard can support using the equation:

m_max = F / g

Where g is the acceleration due to gravity.

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