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3 years ago

9

Which actions are evidence that a chemical reaction has occurred? Select five options. a change in color the formation of a prec

ipitate a clear liquid becoming cloudy a change in shape the formation of bubbles the formation of smoke

1 answer:

vladimir1956 [14]3 years ago

8 0

Answer:

A change in color

The formation of a precipitate

A clear liquid becoming cloudy

The formation of bubbles

The formation of smoke

Explanation:

A chemical reaction can be defined as the conversion of reactant into product. The process in which one or more than one substance is converted into one or many different substances.

There are some indications which can provide the evidence that chemical reaction has occurred.

A change in color of the substance, formation or smoke or bubbles. The formation of the precipitate and in case the substance becomes cloudy.

All these evidences show that chemical reaction has occurred.

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Impact of electricity in the society

Travka [436]

Answer:

<h2>Electricity has many uses in our day to day life. It is used for lighting rooms, working fans and domestic appliances like using electric stoves, A/C and more. All these provide comfort to people. In factories, large machines are worked with the help of electricity.</h2>

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2 years ago

Alternating Current In Europe, the voltage of the alternating current coming through an electrical outlet can be modeled by the

stealth61 [152]

Answer:

$\frac{50}{\pi }$ Hz

Explanation:

In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;

V(t) = V sin (ωt + Ф) -----------------(i)

Where;

V = amplitude value of the voltage

ω = angular frequency = 2 π f [f = cyclic frequency or simply, frequency]

Ф = phase difference between voltage and current.

<u><em>Now,</em></u>

From the question,

V(t) = 230 sin (100t) ---------------(ii)

<em><u>By comparing equations (i) and (ii) the following holds;</u></em>

V = 230

ω = 100

Ф = 0

<em><u>But;</u></em>

ω = 2 π f = 100

2 π f = 100 [divide both sides by 2]

π f = 50

f = $\frac{50}{\pi }$ Hz

Therefore, the frequency of the voltage is $\frac{50}{\pi }$ Hz

7 0

3 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

During the class prize-giving ceremony, Anand clapped his hands hard while Kumar clapped his hands softly. Everybody could hear

algol13

Answer:

C - higher volume

Explanation:

The pitch or frequency of sound that an object can produce depends upon its size and configuration . The shape of hand of all are same so the frequency of sound produced by hands of all will be almost same . Hence frequency of sound produced by the hands of Anand and Kumar would have been almost the same .

But the intensity of sound produced by them would have been different . Intensity represents energy a sound carries . Hard hitting clap will produce sound of higher intensity . Intensity of sound is also called high volume sound . So Kumar's clap will carry greater energy and hence greater volume of sound .

3 0

3 years ago

Peak wavelength of light coming from a star with a temp of 4300k

PSYCHO15rus [73]

from Wein's law

$\frac{k}{ \lambda_{max} } = T \\ \lambda_{max} = \frac{k}{T} \\ \lambda_{max} = \frac{3.898 \times {10}^{ - 3} }{4300} \\ = 9.065 \times {10}^{ - 7}$

3 0

3 years ago