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3 years ago

Through which substance will sound travel the fastest? A)Copper wires B)Sea water C)Hot air D)Cold air

Physics

1 answer:

Papessa [141]3 years ago

3 0

Answer:

A. Copper wires

Explanation:

the the answer is copper wires because sound waves travel the slowest through gases, faster through liquids, and fastest through solids and copper wire is a solid.

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Explain the process of why the balloon is attracted to the wall, and why electrons are not transferred in this process. Is the w

strojnjashka [21]

Answer:

The process by which the balloon is attracted and possibly sticks to the wall is known as static electricity which is the attraction or repulsion between electric charges which are not free to move.

The wall is an insulator.

Explanation:

When a balloon is blown and tied off, and then the balloon is rubbed on the woolly object once in one direction, and the side that was rubbed against the wool is brought near a wall and then released, it is observed that the balloon is attracted to and sticks to the wall. The above observation is due to static electricity.

Static electricity refers to electric charges that are not free to move or that are static. One of the means of generating such charges is by friction. When the balloon is rubbed on the woollen material, electrons are given away to the balloon's surface. Since the balloon is an insulator (materials which do not allow electricity to pass through them easily), the electrons are not free to move. When the balloon is brought near to a wall, there is a rearrangement of the charges present on the wall. Negative charges on the wall move farther away while the positive charges on the wall are attracted to the electrons on the balloon's surface. Because the wall is also an insulator, the charges are not discharged immediately. Therefore, this attraction between opposite charges as well as the static nature of the charges results in the balloon sticking to the wall.

6 0

3 years ago

Electron grops could be considered which of the following

Maurinko [17]

Answer:

Electron groups could be considered as Lone pair electrons and bonded pairs of electrons.

Answer: Option D & B

Explanation:

The two or more electrons can be bonded by single bond, double bond, covalent bond of electrons can simply be lone pair of electrons. Unshared pair of electrons are generally termed as lone pair of electrons in an atom which are generally present in the outermost shell of atoms. Hence electron groups can be determined by bonded pairs and lone pairs of electrons.

I got this from another brainly user

5 0

2 years ago

What are the names of neptunes moons

alexandr402 [8]

Answer:

1. naiad

2.Thalassa

3 Despina

4 Galatea

5 Larissa

6 Hippocamp

7 Proteus ˈ

8 Triton

9 Nereid

10 Halimede

11 Sao

12 Laomedeia

13 Psamathe

14 Neso

Explanation:

hope this helped

5 0

3 years ago

The velocity of particle varies with time as equation

Advocard [28]

Velocity of a particle varies with its displacement as v = ( √(9 ... Velocity of a particle varies with its displacement as v = ( √(9 - x^2) ) m/sFind the magnitude of maximum acceleration of the particle.

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3 0

3 years ago

Read 2 more answers

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

Arturiano [62]

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: