To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,
Q = V*A
Where,
A= Cross-sectional Area
V = Velocity
The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,


Our values are given as,


Re-arrange the equation to find the first ratio of rates we have:



The second ratio of rates is



20 electrons and 2 valence electrons
Answer:
what is your question?..... .
Answer:
The speed when se reaches the top of the incline is 0.28 m/s
Explanation:
The work done is equal to the change of kinetic energy, then:
Wg + Wf + Wn = ΔEk
Where
Wg = work done by gravity
Wf = work done by force
Wn = work done by normal force

Where
m = 8.5 kg
g = 9.8 m/s²
d = 39.93 m
F = 37.4 N
vf = 2 m/s
Replacing:

Answer:
1.997s = 2.0s
Explanation:
From equation of motion,
S = ut + 1/2 gt²
u = 0 m/s
t = ?
g = 9.8 m/s²
S = 1955cm = 19.55m
19.55 = 0 + 1/2 * 9.8 * t²
19.55 * 2 = 9.8t²
t² = 19.55 * 2 / 9.8
t² = 3.987
t = (3.987)^½ (take the square root of both sides)
t = 1.997s = 2.0s