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.
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The energy the electron gives up passing between the electrodes is equal to the product between its charge and the potential difference between the electrodes:

where
e is the electron charge

is the potential difference
Plugging numbers into the equation, we find that the electron gives up is
Any unit of acceleration must have the dimensions (form) of
(a unit of length) / (a unit of time)²
The mass of Jupitar is obtained from the calculations as 5.8 * 10^-14 Kg.
<h3>What is the mass of Jupitar?</h3>
There are nine planets in the solar system and the sun lies at the enter of our solar system. This is the heliocentric model of the solar system.
Given that;
T^2 = GMr^3/4π
T = period
G = gravitational constant
r = radius
M = mass of Jupitar
Now;
1 day = 86400 seconds
1.77 days = 1.77 days * 86400 seconds/1 day
= 152928 seconds
Making M the subject of the formula;
M =4πT^2/Gr^3
M = 4 * 3.142 * (152928)^2/6.67 × 10^-11 * (422 × 10^9)^3
M = 2.9 * 10^11/5.0 * 10^24
M = 5.8 * 10^-14 Kg
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Fnet=fa-fr
=47-23
=24n
W=F∆Xcos°
=24n×0.025cos0
=0.6J