Given:
Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min
Required:
Inlet flowrate
Solution:
The problem can be solved by this general formula.
Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
First, we need to convert the units of the accumulation velocity into m/s to be consistent.
Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s
We then calculate the area of the pool and the area of the orifice by:
Area of pool = 3 × 4 m²
Area of pool = 12m²
Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²
Since we have all we need, we plug in the values to the general equation earlier
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²
Transposing terms,
Inlet flowrate = 0.316 m³/s
Answer:
So they can last longer and have more grip than normal on-road cars. They need that in order for them to run well
Explanation:
That is because work requires energy. According to the law of conservation of energy, it cannot be created or destroyed. When doing work, energy change forms and gets transferred to the object until it is released.
for example, when you lift up an object and place it on a higher elevation, you transferred energy to it and gave it potential energy. The potential energy is transformed into kinetic energy when the object falls down, and if it hits a surface, the energy will scatter, vibrating the areas around it and producing sound.
Also, work= force X distance. The energy does not go away, but rather get changed into some other form of energy
Answer:
It would take the object 5.4 s to reach the ground.
Explanation:
Hi there!
The equation of the height of a free-falling object at any given time, neglecting air resistance, is the following:
h = h0 + v0 · t + 1/2 · g · t²
Where:
h = height of the object at time t.
h0 = initial height.
v0 = initial velocity.
g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).
t = time
Let´s supose that the object is dropped and not thrown so that v0 = 0. Then:
h = h0 + 1/2 · g · t²
We have to find the time at which h = 0:
0 = 470 ft - 1/2 · 32.2 ft/s² · t²
Solving for t:
-470 ft = -16.1 ft/s² · t²
-470 ft / -16.1 ft/s² = t²
t = 5.4 s
