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3 years ago

Which is not a common property of ionic compounds

Physics

1 answer:

Naddik [55]3 years ago

6 0

<h2>Answer:</h2>

Low melting points and electrical conductivity in solids are not common properties of ionic solids.

<h3>Explanation:</h3>

In ionic compounds the electrons involved in the bonding are tightly packed under the influence of electrostatic force of attraction. So the movement of these electrons is very difficult.

In the melting point the bond breaking between the atoms is involved. Hence in case of ionic compound there is a high amount of energy needed to break the ionic bonds.

Electrical conductivity involves the free movement of electrons which is impossible in ionic solids.

So low melting points and electrical conductivity in solids are not common properties of ionic solids.

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Which one of the following types of electromagnetic wave travels through space the fastest?

Setler [38]

Answer:

As a result, light travels fastest in empty space, and travels slowest in solids. In glass, for example, light travels about 197,000 km/s.

Explanation:

4 0

2 years ago

Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0

3 years ago

Choose 3 ways to keep our bodies healthy

Alina [70]

Answer:

take daily showers . eat vegies .sanitize your hands

7 0

2 years ago

Read 2 more answers

The measure of a spring’s resistance to being compressed or stretched is the

larisa86 [58]

<u>Answer;</u>

<em>Spring constant </em>

<u>Explanation;</u>

The measure of a spring’s resistance to being compressed or stretched is the <u>spring constant</u>.

The symbol of spring constant is K, since it is a constant. From the Hooke's law,for a helical spring or any elastic material, the extension force is directly proportional to the extension provided the elastic limit is not exceeded.

Therefore; the spring constant = Force/extension. That is; K = F/e; where k is the spring constant, F is the extension force and e is the extension.

Spring constant depicts the resistance of the spring to compressional and stretching forces.

7 0

3 years ago

Read 2 more answers

Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is

kobusy [5.1K]

Answer:

The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

$Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}$

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

$Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa$

Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

6 0

3 years ago