Question

Appreciation of poem the voice of the rain
grade 11

Answer 1

The Voice of The Rain Central Idea of the Poem

The poem The Voice of the Rain’ by Walt Whitman signifies the eternal role that the rain plays in nurturing, quenching and purifying the various elements of Earth. The rain returns the favour to its place of origin from where it rises unseen from the depths of the water and from the land. The rain itself is explaining to the reader about its origin, work and its cyclic movement. A comparison has also been drawn between rain and music as both of them make the world more lively and return to their place of origin after fulfilling their purpose.

Answer 2

I'd give you a call at your earliest convenience to

$</p><p>\underline{\bf{Given\::}} </p><p>Given:</p><p> </p><p> </p><p></p><p></p><p></p><p>\underline{\bf{To\:find\::}} </p><p>Tofind:</p><p> </p><p> </p><p></p><p></p><p>\underline{\bf{Explanation\::}} </p><p>Explanation:</p><p> </p><p> </p><p></p><p></p><p>\boxed{\bf{\frac{1}{f} =\frac{1}{v} -\frac{1}{u} }}}}</p><p></p><p>\begin{gathered}\longrightarrow\sf{\dfrac{1}{-10} =\dfrac{1}{v} -\dfrac{1}{-30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-10} +\dfrac{1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{-3+1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\cancel{\dfrac{-2}{30} }}\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-15} }\\\\\\\longrightarrow\sf{v=-15\:cm}\end{gathered} </p><p>⟶ </p><p>−10</p><p>1</p><p> </p><p> = </p><p>v</p><p>1</p><p> </p><p> − </p><p>−30</p><p>1</p><p> </p><p> </p><p>⟶ </p><p>v</p><p>1</p><p> </p><p> = </p><p>−10</p><p>1</p><p> </p><p> + </p><p>30</p><p>1</p><p> </p><p> </p><p>⟶ </p><p>v</p><p>1</p><p> </p><p> = </p><p>30</p><p>−3+1</p><p> </p><p> </p><p>⟶ </p><p>v</p><p>1</p><p> </p><p> = </p><p>30</p><p>−2</p><p> </p><p> </p><p> </p><p> </p><p>⟶ </p><p>v</p><p>1</p><p> </p><p> = </p><p>−15</p><p>1</p><p> </p><p> </p><p>⟶v=−15cm</p><p> </p><p> </p><p></p><p></p><p>\boxed{\bf{M \:A \:G \:N\: I \:F \:I \:C\: A\: T \:I \:O\: N :}} </p><p>MAGNIFICATION:</p><p> </p><p> </p><p></p><p>\begin{gathered}\mapsto\sf{m=\dfrac{Height\:of\:image\:(I)}{Height\:of\:object\:(O)} =\dfrac{Distance\:of\:image}{Distance\:of\:object} =\dfrac{v}{u} }\\\\\\\mapsto\sf{m=\cancel{\dfrac{-30}{-15}} }\\\\\\\mapsto\bf{m=2\:cm}\end{gathered} </p><p>↦m= </p><p>Heightofobject(O)</p><p>Heightofimage(I)</p><p> </p><p> = </p><p>Distanceofobject</p><p>Distanceofimage</p><p> </p><p> = </p><p>u</p><p>v</p><p> </p><p> </p><p>↦m= </p><p>−15</p><p>−30</p><p> </p><p> </p><p> </p><p> </p><p>↦m=2cm</p><p> </p><p> </p><p></p><p>Thus;</p><p></p><p>The magnification will be 2 cm .</p><p></p><p>$