To calculate the time it takes to purge a 10-mm section of a 20-cm diameter pipe, first find the volume of the section. The volume can be calculated using the formula for the volume of a cylinder, ( V = \pi r^2 h ), where ( r ) is the radius (0.1 m) and ( h ) is the length (0.01 m). This results in a volume of approximately 0.000314 m³ or 0.314 L. At a flow rate of 17 L/min, it would take about 0.0185 minutes, or approximately 1.11 seconds, to purge the section.
The removal of the tumor would effectively restore the trachea's diameter and consequently increase the area of its cross section. This would lead to an improvement in the flow of air through the trachea due to reduced obstruction.
Approximately 33.1 inches. If you need it more precise: 33.104256125083159 inches.
The field of view would be 80 times the diameter.
The bolometric correction allows you to convert between visual and bolometric (total) magnitude - where the bolometric magnitude includes all radiation emitted by the star, not just visible light. It has nothing to do with the angular diameter.
To find the size of the cell, you can divide the diameter of the field of vision by 13. Given that the diameter of the field of vision at high power is 600 micrometers, the calculation would be: 600 micrometers / 13 ≈ 46.15 micrometers. Therefore, the size of the cell is approximately 46.15 micrometers.
Area equal pi * r^2. Diameter of six equals radius of three. So according to Google, 9 * pi * (15*12) = 5089 cubic inches. Now, again according to Google, 35 (ft^3) = 60 480 in^3 per hour. I'm going to leave you to figure out the answer from here!
The diameter is twice the radius - so, using your example, the diameter would be 6.
The removal of the tumor would effectively restore the trachea's diameter and consequently increase the area of its cross section. This would lead to an improvement in the flow of air through the trachea due to reduced obstruction.
My $210 Physical Geography class text book does not have the answer. It just says it would be smaller than the diameter of a cross section through the equator.
0.000095meters = 0.095mm Dia of Hair = 0.095 + 10 = 10.095mm Diameter = 10.095mm
The diameter of 64 inches is simply 64 inches. If you are referring to the diameter of a circle with a circumference of 64 inches, you can calculate it using the formula diameter = circumference/π. In that case, the diameter would be approximately 20.37 inches.
I'm asking the exact same question...you go to Mohawk College don't you?
Approximately 33.1 inches. If you need it more precise: 33.104256125083159 inches.
The circumference of a circle can be calculated using the formula ( C = \pi \times d ), where ( d ) is the diameter. For a diameter of 1.75, the circumference would be ( C = \pi \times 1.75 ), which is approximately 5.49 when using ( \pi \approx 3.14 ). Therefore, the circumference of a circle with a diameter of 1.75 is about 5.49 units.
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The circumference of a circle can be calculated using the formula ( C = \pi \times d ), where ( d ) is the diameter. If the diameter is 4 feet, the circumference would be ( C = \pi \times 4 ) feet, which is approximately 12.57 feet when using ( \pi \approx 3.14 ).
Yes, and No, Nitrogen can be used to purge gas lines prior to welding, but the correct answer depends on several things, mainly, your welding application. A significant amount of resistance welding applications use N2 as a suitable environment for welding. Other applications use N2 as a purge gas, and then purge the N2 with the process gas. That said, if you were going to be using a different inert gas (Argon, Helium, Xenon, etc.) to provide the actual the actual shield for the weld, then you would still need to displace (purge) the N2 from the lines. The last statement leads to the question of; if you are going to have to purge the lines anyway, why do it twice? The answer to this question is pretty basic. Nitrogen is cheap. If you had to purge a lot of lines, or a large volume of space, you would rather use N2 to drive out the Oxygen and lower the dew point (moisture). Using one of the other welding gases would cost about ten times as much to perform the operation.