Question

The distance between the receiver and the sender stations is 200 km. Find the letters between ground waves and whirlwinds from the time the sender starts reaching Aakhai. Given that, the practice of anemia: the height of the surface is about 100 km. ​

Answer 1

$\underline{\underline{ \large{ \bf \: Answer-}}} \\ \\= 2.76× 10^{-4}$

$\underline{\underline{ \bf{ \huge{Solution- }}}}$

$\underline{ \underline{ \large{ \bf \: Given - }}}$

According to the question we know that

The distance between the receiver and the sender station is 200km, and The hight of surface is about 100km

$\large{\underline{ \underline{ \mathbf{Need \: to \: find -}}}}$

The letters between ground waves and whirlwinds from the time the sender starts reaching Aakhai.

$\underline{ \underline{ \large{\bf \:Uses \: Formulas - \:} }}$

1. Movement of electromagnetic waves c = 3x10^8m / sec

2. BC = √(OC)² + (OB)²

3. Time = Speed / c

$\underline{\underline{\bf \huge{Explanation-}}}$

We know the movement of electrical magnetic waves is 3 × 10^8 m/s

we know Soil wave moves from the sender to the Soyo line

$\bf \: So, \: the \: summer \: vapor \: waves \: taken \: by \: \\ \bf ground \: waves \: that \: reach$

$t _{g} = \frac{ \bf \: speed \: AB}{c} \:$

$\bf{Let \: put \: the \: above \: values}$

$\implies \: \bf {\: \frac{200 \times {10}^{3} \: m }{3 \times {10}^{8} \frac{m}{s} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large{\{ } \bf{Given \:} \\ \\ \\ \implies \: \: \boxed{ \bf{6.67 \times {10}^{ - 4} \: sec }}$

The spirits come to the first ionosphere, then return to the morning after reflecting.

The height of the ionosphere

$\mathbf {\: h = OC = 100 \: km} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge{\{} \small{ \bf \: Given }$

$\bf \: OB \: = 100 \: km$

$so \: \: \: \bf \: AC = BC = \sqrt{OC {}^{2} + OB {}^{2} } \\ \\ \\ \implies \bf \: \sqrt{ {(100)}^{2} + {(100)}^{2} } = \sqrt[100]{2} \: \: \: km$

$\bf The \: distance \: trave l \: by \: spirits \: waves$

$\mathbf{ = AC + CB = 2AC \: = \sqrt[200]{2} \: km} \: \: \: \: \huge \{ \small\: we \: know \: that \: AC = BC \: \\ \\ \\ \implies \: \: t _{s} = \bf \frac{2AC}{c} \\ \\ \\ \implies \bf{ \frac{ \sqrt[200]{2} \times {10}^{3} m}{3 \times {10}^{8} \frac{m}{s} } } \\ \\ \\ \implies \bf{6.67 \sqrt{2} \times {10}^{ - 4} } \\ \\ \\ \implies \bf 9.43 \times {10}^{ - 4} sec \\ \\ \\ \implies \: \therefore \: \: \: \Delta \: t \: = t _{s} - t _{g} \\ \\ \\ \implies \: \: \: \boxed{ \bf \: \: 2.76 \times {10}^{ - 4} }$