Answer:
"2Ω" is the net resistance in the circuit.
Explanation:
The given resistors are:
R1 = 3Ω
R2 = 6Ω
The net resistance will be:
⇒ ![\frac{1}{R_{net}} =\frac{1}{R_1} +\frac{1}{R_2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7Bnet%7D%7D%20%3D%5Cfrac%7B1%7D%7BR_1%7D%20%2B%5Cfrac%7B1%7D%7BR_2%7D)
On substituting the values, we get
⇒ ![\frac{1}{R_{net}} =\frac{1}{3} +\frac{1}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7Bnet%7D%7D%20%3D%5Cfrac%7B1%7D%7B3%7D%20%2B%5Cfrac%7B1%7D%7B6%7D)
On taking L.C.M, we get
⇒ ![\frac{1}{R_{net}} =\frac{2+1}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7Bnet%7D%7D%20%3D%5Cfrac%7B2%2B1%7D%7B6%7D)
⇒ ![\frac{1}{R_{net}} =\frac{3}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7Bnet%7D%7D%20%3D%5Cfrac%7B3%7D%7B6%7D)
⇒ ![\frac{1}{R_{net}} =\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7Bnet%7D%7D%20%3D%5Cfrac%7B1%7D%7B2%7D)
On applying cross-multiplication, we get
⇒ ![R_{net}=2 \Omega](https://tex.z-dn.net/?f=R_%7Bnet%7D%3D2%20%5COmega)
Answer:
In physics the standard unit of weight is Newton, and the standard unit of mass is the kilogram. On Earth, a 1 kg object weighs 9.8 N, so to find the weight of an object in N simply multiply the mass by 9.8 N. Or, to find the mass in kg, divide the weight by 9.8 N.
Explanation:
<em><u>Radhe</u></em><em><u> </u></em><em><u>Radhe</u></em><em><u>❤</u></em>
C) In the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.
hope this helps and have a great day :)
Answer:
The relationship is only between the coefficients A, E and J which is:
. The remaining coefficients can be anything without any constraints.
Explanation:
Given:
The three components of velocity is a velocity field are given as:
![u = Ax + By + Cz\\\\v = Dx + Ey + Fz\\\\w = Gx + Hy + Jz](https://tex.z-dn.net/?f=u%20%3D%20Ax%20%2B%20By%20%2B%20Cz%5C%5C%5C%5Cv%20%3D%20Dx%20%2B%20Ey%20%2B%20Fz%5C%5C%5C%5Cw%20%3D%20Gx%20%2B%20Hy%20%2B%20Jz)
The fluid is incompressible.
We know that, for an incompressible fluid flow, the sum of the partial derivatives of each component relative to its direction is always 0. Therefore,
![\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D%2B%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D%2B%5Cfrac%7B%5Cpartial%20w%7D%7B%5Cpartial%20z%7D%3D0)
Now, let us find the partial derivative of each component.
![\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(Ax+By+Cz)\\\\\frac{\partial u}{\partial x}=A+0+0=A\\\\\frac{\partial v}{\partial y}=\frac{\partial }{\partial y}(Dx+Ey+Fz)\\\\\frac{\partial v}{\partial y}=0+E+0=E\\\\\frac{\partial w}{\partial z}=\frac{\partial }{\partial z}(Gx+Hy+Jz)\\\\\frac{\partial w}{\partial z}=0+0+J=J](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D%3D%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D%28Ax%2BBy%2BCz%29%5C%5C%5C%5C%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D%3DA%2B0%2B0%3DA%5C%5C%5C%5C%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D%3D%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D%28Dx%2BEy%2BFz%29%5C%5C%5C%5C%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D%3D0%2BE%2B0%3DE%5C%5C%5C%5C%5Cfrac%7B%5Cpartial%20w%7D%7B%5Cpartial%20z%7D%3D%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20z%7D%28Gx%2BHy%2BJz%29%5C%5C%5C%5C%5Cfrac%7B%5Cpartial%20w%7D%7B%5Cpartial%20z%7D%3D0%2B0%2BJ%3DJ)
Hence, the relationship between the coefficients is:
![A+E+J=0](https://tex.z-dn.net/?f=A%2BE%2BJ%3D0)
There is no such constraints on other coefficients. So, we can choose any value for the remaining coefficients B, C, D, F, G and H.
Answer:
Your answer should be Cooled Air
Explanation: