The equal velocity approach for duct size assumes that the air velocity in each duct segment is the same.
How fast is the air moving through a duct?
The most common unit of air velocity (distance traveled in a unit of time) is feet per minute (FPM). The amount of air passing past a location in the duct per period of time may be calculated by multiplying the airflow by the area of the duct. The standard unit for volume flow is cubic feet per minute (CFM).
What happens when the size of ducts changes to the airflow?
- Result for an image The equal velocity technique for duct size makes the assumption that air velocity is constant across the entire duct system.
- The main lesson to be learned from this is that when air goes from a bigger to a narrower duct, its velocity rises. The velocity drops when it transitions from a shorter to a bigger duct. The flow rate or the amount of air passing through the duct in cubic feet per minute is the same in all scenarios.
Learn more about air velocity here:
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Answer:
C) If an ice cube is placed into a boiling water, then it will melt in less than 2 minutes.
Explanation:
Answer: It will increase the length of the day
Explanation: The polar ice caps contain the solid ice particles found around the polar regions,an increased temperature is a major cause of the melting of the polar ice Caps. When the polar ice caps melt especially in huge volumes they increase the amount of water flowing into the Sea this will cause a rise in the volume of the sea.
This rise in sea level will add some weights to Earth, making the Earth to rotate slightly lower than its normal speed,a slower rotation will lead to longer days.
Answer:
sorry but i cannot understand your language, can you speak in english
Momentum will be conserved in one dimension in the explosion.
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Given that the fragment a acquires three
times the kinetic energy of the fragment b.
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P</span><span><span>initial </span><span>= p</span></span>final ⇒ 0 =mₐv⁰ₐ+mьv⁰ь= 0 ⇒ v⁰ь = -mₐv⁰ₐ/mь
KE= 3KEь
⇒1/2 mₐv⁰ₐ² = 3 (1/2mьv⁰ь²)
</span><span>
⇒1/2 mₐv⁰ₐ² = 3/2 mь(-mₐv⁰ₐ/mь)²
⇒1/2 mₐv⁰ₐ² = 3/2 mь(mₐ²v⁰ₐ²/mь²)
</span>
⇒1/2 x 2/3 = mₐ/mь= 1/3
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<span>
Thus the ratio
of the masses of the fragments is 1:3.
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