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2 years ago

Use the same line to answer the questions

Physics

1 answer:

vichka [17]2 years ago

7 0

(a) The distance From A to B to M to N is 22 m, the displacement is 22 m, the speed is 4.4 m/s and the average velocity is 4.4 m/s.

(b) The distance From C to A to G to D is 24 m, the displacement is 2 m, the speed is 4.8 m/s and the average velocity is 0.4 m/s.

(c) The distance From J to I to E to H is 16 m, the displacement is -2 m, the speed is 3.2 m/s and the average velocity is -0.4 m/s.

(d) The distance From F to B to K to F is 32 m, the displacement is 0 m, the speed is 6.4 m/s and the average velocity is 0 m/s.

<h3>Distance traveled by the object</h3>

The distance traveled by the object is calculated as follows;

Distance From A to B to M to N

From A to B to M to N = 3 + 18 + 1 = 22

Distance From C to A to G to D

= 5 + 13 + 6

= 24

Distance From J to I to E to H

= 1 + 8 + 7

= 16

Distance from F to B to K to F

= 8 + 16 + 8

= 32

<h3>Displacement of the object</h3>

The displacement of the object is calculated as follows;

displacement = final position - initial position

Displacement From A to B to M to N

= 11 - (-11)

= 22

Displacement from C to A to G to D

= D - C

= - 4 - (-6)

= 2

Displacement from J to I to E to H

= H - J

= 5 - 7

= - 2

Displacement From F to B to K to F

= F - F

= 0

<h3>Speed of the object </h3>

The speed of the object is calculated as follows;

speed = total distance/total time

Speed of the object from A to B to M to N

= 22/5 = 4.4 m/s

Speed of the object from C to A to G to D

= 24/5

= 4.8 m/s

Speed of the object from J to I to E to H

= 16/5

= 3.2 m/s

Speed of the object from F to B to K to F

= 32/5

= 6.4 m/s

<h3>Average velocity of the object </h3>

The average velocity of the object is calculated as follows;

Average velocity from A to B to M to N

= 22/5

= 4.4 m/s

Average velocity of the object from C to A to G to D

= 2/5

= 0.4 m/s

Average velocity of the object from J to I to E to H

= -2/5

= -0.4 m/s

Average velocity of the object from F to B to K to F

= 0

Learn more about average velocity here: brainly.com/question/24445340

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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: