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A Mosquito Net Over a 7 Ft × 4 Ft Bed is 3 Ft High. the Net Has a Hole at One Corner of the Bed Through Which a Mosquito Enters the Net. It Flies and Sits at the Diagonally Opposite - Physics

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प्रश्न

A mosquito net over a 7 ft × 4 ft bed is 3 ft high. The net has a hole at one corner of the bed through which a mosquito enters the net. It flies and sits at the diagonally opposite upper corner of the net. (a) Find the magnitude of the displacement of the mosquito. (b) Taking the hole as the origin, the length of the bed as the X-axis, it width as the Y axis, and vertically up as the Z-axis, write the components of the displacement vector.

संक्षेप में उत्तर
योग
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उत्तर

Displacement vector of the mosquito,

\[\vec{r} = 7 \hat {i} + 4 \hat {i} + 3 \hat {k}\]
(a) Magnitude of displacement
\[= \sqrt{7^2 + 4^2 + 3^2}\]
\[= \sqrt{74} \text { ft }\]
(b) The components of the displacement vector are 7 ft, 4 ft and 3 ft along the X, Y and Z-axes, respectively.
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अध्याय 2: Physics and Mathematics - Exercise [पृष्ठ २९]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
अध्याय 2 Physics and Mathematics
Exercise | Q 9 | पृष्ठ २९

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