Answer:
V = 6.36 m³
Explanation:
For this exercise we will use fluid mechanics relations, starting with the continuity equation.
Let's write the flow equation
Q = v₁ A₁
The area of a circle is
A = π r²
Radius is half the diameter
A = π/4 d²
Q = v₁ π/4 d₁²
Q = π/ 4 15 0.03 2
Q = 0.0106 m3 / s
The volume of water in t = 10 min = 10 60 = 600 s
Q = V / t
V = Q t
V = 0.0106 600
V = 6.36 m³
The weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
To find the answer, we have to know about the pressure.
<h3>How to find the weight of a column of air?</h3>
- As we know that the expression of pressure as,

where; F is the force, here it is equal to the weight of the air column, and A is the area of cross section.
- It is given that, the air column is extending from earth's surface to the top of the atmosphere, thus, the pressure will be atmospheric pressure,

- From this, the value of weight will be,

Thus, we can conclude that, the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
Learn more about the pressure here:
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Answer:
Explanation:
Electrical energy is:
E = IVt
E = 210 * 12 * 10\\\\\\E = 25200 J
Oil flows upward in the wick of a lantern because of the liquid property called <span>Capillarity c:</span>
Answer:
The answer is <em>15A</em>
Explanation:
Terminal of like polarity are connected with jumpers of negligible resistance
Kirchoff's voltage law gives
E1 -E2 - IR1 - IR2 = 0
I(R1 + R20 = E1 -E2
I= 12 -9/ 0.18 + O.2 = 15A OF CURRENT WILL FLOW THROUGH DISCHARGED BATTERY.