Answer:
different value to l if n=5 are 0,1,2,3,4
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:
5/9
Explanation:
First, find a common factor for 30 and 54.
30 and 54 can both be divided by 2.
30 divided by 2 = 15
54 divided by 2 = 27
So, your new fraction is 15/27.
However, don't stop there.
15 and 27 can both be divided by 3 as well.
15 divided by 3 = 5
27 divided by 3 = 9
So, your final answer is 5/9.
I hope this helps! :)
This is honestly not something I’ve learned. The answer is something I don’t know.
Answer:
2, 8 and shell
Explanation:
Neon as atomic number 10. Since for each shell, electrons equal 2n².
When n = 1, 2n² = 2(1)² = 2
When n = 2, 2n² = 2(2)² = 8
So it fills both the first and second shell with 2 and 8 electrons respectively to achieve its stable atomic state. The rest of the 8 electrons go into the second shell because the first shell has achieved its stable dual configuration of two electrons. The next shell requires a maximum of 8 electrons to achieve stability so, the remaining electrons fill it up to achieve the stable octet configuration.
