Recent content by Aurelius120

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    Oil enters a bend in a pipe -- calculate the force needed to steady bend

    ##P_2A_2 cos37##? Since force perpendicular to area is taken?
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    Oil enters a bend in a pipe -- calculate the force needed to steady bend

    The fluid flows horizontally in ##+\hat i## upto bend so force on it must be in direction of its velocity to make it flow and its reaction force on pipe must be in ##-\hat i## After the bend the fluid flows in ##-cos37\hat i+sin37\hat j## and reaction force will be in opposite direction
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    Oil enters a bend in a pipe -- calculate the force needed to steady bend

    Reaction forces due to oil pressure: $$\vec F_1=P_1A_1 \hat i=-56\times 10^3\hat i$$ $$\vec F_2=P_2A_2=(8.6\times 10^3) \cos37\hat i-(8.6\times 10^3)\sin 37\hat j$$ Net force on pipe =Sum of Reaction force due to oil pressure= $$|\vec F_1+\vec F_2|=\sqrt{(56000-860\times 8)^2+(860\times...
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    Oil enters a bend in a pipe -- calculate the force needed to steady bend

    The pressure is easily calculated from equation of continuity and Bernoullis theorem: $$A_1v_1=A_2v_2\implies v_2=16ms^{-1}$$ SInce pipe is in horizontal plane, no difference in pressure because of height $$P_1+\frac{\rho v_1^2}{2}=P_2+\frac{\rho v_2^2}{2}$$ $$P_2=172\times 10^3$$ What I...
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    Equation of hyperbola confocal with ellipse having same principal axes

    Also just to clarify more than one options are correct so all need to be checked
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    Equation of hyperbola confocal with ellipse having same principal axes

    The equation of ellipse reduces to : $$(2x+3)^2+(3y+2)^2=8$$ $$\frac{(x+3/2)^2}{8/4}+\frac{(y+2/3)^2}{8/9}=1$$ Center of ellipse =##\left(\frac{-3}{2},\frac{-2}{3}\right)## ##b^2=a^2(1-e^2)=8/9## and ##a^2=8/4## Therefore ##e=\frac{\sqrt{5}}{3}## Distance between foci=##\frac{2\sqrt{10}}{3}##...
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    Electric field & force due to charged insulating hemispherical shells

    The book's answer is option C though and the solution on the web is as described in the 1st post by somehow using the formula of infinite sheet
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    Electric field & force due to charged insulating hemispherical shells

    So I know I have to equate force on a hemispherical shell with spring force to get value of compression but I can't find the force on the hemispheres Some places that do have the solution use the formula : $$\text{Field of non-conducting hemispherical shell= } \frac{\sigma}{2\epsilon_○} $$ This...
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    A vertical U tube rotates about the vertical axis -- Find omega to cause this height offset

    But isn't it equating the force required for rotation to force exerted? I don't know if there is something like Centripetal pressure that is necessary for rotation
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    A vertical U tube rotates about the vertical axis -- Find omega to cause this height offset

    I am not sure I understand. The water can't be treated as rectangular/cylindrical blocks with forces ##V\rho g## on either side and arbitrary force due pressure in-between❓️
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    A vertical U tube rotates about the vertical axis -- Find omega to cause this height offset

    I think I got it $$\int^b_0{\omega^2 A\rho rdr}-PA=A(b)\rho g$$ $$\int^{2b}_0{\omega^2A\rho rdr}-PA=A(3b)\rho g$$ Therefore Subtracting equation 1 from 2 $$\frac{3A\rho\omega^2b^2}{2}=2Ab\rho g$$
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    A vertical U tube rotates about the vertical axis -- Find omega to cause this height offset

    So how do I account for that? Its arbitrary so how to know its value
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    A vertical U tube rotates about the vertical axis -- Find omega to cause this height offset

    So it will be $$\int{\omega^2r dm}= V\rho g$$ On either side with limits ##0\text{ to }b## and ##0\text{ to }2b## and ##V## changed according to height of liquid column
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    A vertical U tube rotates about the vertical axis -- Find omega to cause this height offset

    So I have to integrate or something? Isn't the pressure atmospheric everywhere , so it doesn't affect the net results anywhere?
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