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2 years ago

HEY CAN ANYONE PLS HELP ME OUT IN DIS PLS!

Physics

2 answers:

Leona [35]2 years ago

8 0

The significant figure rules for addition and subtraction basically has you add or subtract as normal, but the rounding is where the rule kicks in. Specifically you round based on the least accurate decimal portion (the term with the least number of digits to the right of the decimal point).

We do this because we want to make sure we don't involve digits we aren't 100% sure of their accuracy.

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Part A

Add up the values like normal

32.567+135.0+1.4567 = 169.0237

Now we look at the decimal portions of each number. We look at the number of digits to the right of each decimal point.

32.567 has three such digits (5,6 and 7)

135.0 only has one digit (the 0)

1.4567 has four digits (4,5,6,7)

The least accurate is 135.0 since we only know this to the tenths place. So we'll be rounding the sum we got above (169.0237) to the nearest tenth

169.0237 rounds to 169.0

<h3>Answer before rounding = 169.0237</h3><h3>Answer after rounding = 169.0</h3>

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Part B

Add like normal

246.24+238.278+98.3 = 582.818

Then round to one decimal place since this is the fewest number of digits after the decimal point (in the number 98.3)

582.818 rounds to 582.8

<h3>Answer before rounding = 582.818</h3><h3>Answer after rounding = 582.8</h3>

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Part C

Subtraction is handled in a similar fashion

4.829-0.26 = 4.569

Which rounds to 4.57 since two digits is the smallest number of digits after the decimal point. Put another way, 0.26 is the least of the accurate measurements comparing 4.829 and 0.26; we always go with the least accurate or weakest link when it comes to rounding sig figs. If we went the other way around, then we could be introducing data that isn't really there.

<h3>Answer before rounding = 4.569</h3><h3>Answer after rounding = 4.57</h3>

ryzh [129]2 years ago

3 0

Answer:

A) 169.0167

Explanation:

ive done this before

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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: