As per IS 456 : 2000, the maximum permissible shear stress, τc,max, is based on Diagonal tension failure Diagonal compression failure Flexural tension failure Flexural compression failure

Answer:

23 A

Explanation:

consider the above part of circuit as shown in figure and applying Kirchhoff's current law we obtain

I

1

+5−I

2

=0

I

1

=I

2

−5..................................................(1)

I

2

+15−I

3

=0

I

2

=I

3

−15..........................................(2)

I

3

+3−I

4

=0

I

3

=I

4

−3..........................................(3)

and I

4

−i−I

1

=0

I

4

=I

1

+i...............................................(4)

now using values from equations(2), (3) and (4) into (1)we get

I

1

=I

1

+i−23

∴ i= 23 A

Answer:

[tex]\qquad\boxed{ \sf{ \: \bf \:Length \: of \: Latus \: Rectum \: = \: \dfrac{28}{13} \: }} \\ \\ [/tex]

Step-by-step explanation:

Given equation of parabola is

[tex]\sf \:169[ {(x - 1)}^{2} + {(y - 3)}^{2}] = {(5x - 12y + 17)}^{2} \\ \\ [/tex]

From this definition of parabola, we have

[tex]\sf \:Focus \: of \: parabola \: = \: (1, \: 3) \\ \\ [/tex]

and

[tex]\sf \:Equation \: of \: directrix \:is \:5x - 12y + 17 = 0 \\ \\ [/tex]

We know,

Length of Latus Rectum of parabola = 2 × length of perpendicular drawn from focus on directrix.

i.e

Length of Latus Rectum of parabola = 2 × length of perpendicular drawn from (1, 3) on 5x - 12y + 17 = 0.

So,

[tex]\sf \:Length \: of \: Latus \: Rectum \: = \: 2 \times \bigg |\dfrac{5(1) - 12(3) + 17}{ \sqrt{ {5}^{2} + {( - 12)}^{2} } } \bigg| \\ \\ [/tex]

[tex]\sf \:Length \: of \: Latus \: Rectum \: = \: 2 \times \bigg |\dfrac{5 -36 + 17}{ \sqrt{ 25 + 144 } } \bigg| \\ \\ [/tex]

[tex]\sf \:Length \: of \: Latus \: Rectum \: = \: 2 \times \bigg |\dfrac{ -14}{ \sqrt{169} } \bigg| \\ \\ [/tex]

[tex]\sf \:Length \: of \: Latus \: Rectum \: = \: 2 \times \dfrac{14}{13} \\ \\ [/tex]

[tex]\sf\implies \bf \:Length \: of \: Latus \: Rectum \: = \: \dfrac{28}{13} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf {y}^{2} = 4ax & \bf \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf Vertex & \sf (0,0) \\ \\ \sf Focus & \sf (a,0) \\ \\ \sf Length \: of \: latus \: rectum & \sf 4a\\ \\ \sf Equation \: of \: axis & \sf y = 0\\ \\ \sf Equation \: of \: directrix & \sf y = - a \end{array}} \\ \end{gathered} \\ [/tex]

[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf {y}^{2} = - 4ax & \bf \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf Vertex & \sf (0,0) \\ \\ \sf Focus & \sf ( - a,0) \\ \\ \sf Length \: of \: latus \: rectum & \sf 4a\\ \\ \sf Equation \: of \: axis & \sf y = 0\\ \\ \sf Equation \: of \: directrix & \sf y = a \end{array}} \\ \end{gathered} \\ [/tex]

[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf {x}^{2} = 4ay & \bf \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf Vertex & \sf (0,0) \\ \\ \sf Focus & \sf (0,a) \\ \\ \sf Length \: of \: latus \: rectum & \sf 4a\\ \\ \sf Equation \: of \: axis & \sf x = 0\\ \\ \sf Equation \: of \: directrix & \sf x = - a \end{array}} \\ \end{gathered} \\ [/tex]

[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf {x}^{2} = \: - \: 4ay & \bf \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf Vertex & \sf (0,0) \\ \\ \sf Focus & \sf (0, - a) \\ \\ \sf Length \: of \: latus \: rectum & \sf 4a\\ \\ \sf Equation \: of \: axis & \sf x = 0\\ \\ \sf Equation \: of \: directrix & \sf x = a \end{array}} \\ \end{gathered} \\ [/tex]

Answer:

188.4m

Explanation:

By applying E=Asin (+Kx+omega.t) .

The lanthanide that would show colour is:

Gd3+

La3+

Sm3+