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3 years ago

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You are lying on a beach, your eyes 20cm above the sand. Just as the Sun sets, fully disappearing over the horizon, you immediat

ely jump up, your eyes now 150 cm above the sand, and you can again just see the top of the Sun. If you count the number of seconds until the Sun fully disappears again, you can estimate the Earth’s radius. But for this Problem, use the known radius of the Earth to calculate the time t.

1 answer:

Naya [18.7K]3 years ago

6 0

Answer:

The time is $t = 8.78 \ s$

Explanation:

From the question we are told that

The initial height of the eye is $h _1 = 20 \ cm = 0.2 \ m$

The height of the eye when you jumped up is $h_2 = 150 \ cm = 1.5 \ m$

An illustration of this question is shown on the first uploaded image

Generally the radius of the earth is $R = 6.38*10^{6} \ m$

Now from the diagram first sun means first time you saw the sun and the second sun means second time you saw the sun then

H is the height increase when you quickly stood up which is mathematically evaluated as

$H = h_2 -h_1$

$H = 1.5 - 0.2$

$H = 1.3 \ m$

Also $\theta$ i the angular displacement between the first and second position and from geometry it is also the angle at one of the sides of the right angle triangle

Applying Pythagoras theorem

$(R+H)^2 = K^2 + R^2$

=> $R^2 + H^2 + 2RH = K^2 + R^2$

Now given that H is very small compared to R the we ignore $H^2$

So

$R^2 + 2RH = K^2 + R^2$

=> $K = \sqrt{2RH}$

=> $K = \sqrt{2 * 6.38*10^6 * 1.3 }$

=> $K = 4073 \ m$

Now the $\theta$ is mathematically evaluated using SOHCAHTOA as follows

$tan \theta = \frac{K}{R}$

$\theta = tan^{-1}[ \frac{K}{R}]$

=> $\theta = tan^{-1}[ \frac{ 4073}{6.38*10^{6}}]$

=> $\theta = 0.0366^o$

Generally

$1 \revolution\ around \ the\ earth = 24 \ hours = 86400 \ seconds = 360 ^o$

So

$\frac{\theta}{360} = \frac{t}{86400}$

=> $\frac{0.0366}{360} = \frac{t}{86400}$

=> $t = 8.78 \ s$

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3 years ago

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

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Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

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