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3 years ago

Describe what happens in a chemical reaction

Physics

2 answers:

ollegr [7]3 years ago

7 0

Answer:

LA REPUESTA CORRECTA ES

Explanation:

Lyrx [107]3 years ago

3 0

Chemical reaction, a process in which one or more substances, the reactants, are converted to one or more different substances, the products. Substances are either chemical elements or compounds. A chemical reaction rearranges the constituent atoms of the reactants to create different substances as products

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Properties of convex mirror

Ahat [919]

Answer:

The image result of an object reflected by a convex mirror is typically virtual, upright, and smaller. Discover how moving the object farther away from the mirror's surface affects the size of the virtual image formed behind the mirror

Explanation:

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3 years ago

Which of the following are correct statements about the way an atom is put

Ann [662]

Answer:

valenc e shell

Explanation:

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3 years ago

A. Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with

zzz [600]

(a) 10 GHz is the frequency of microwave radiation.

(b) 0.167 ms is required by the microwave to travel between two mountains.

Answer:

Explanation:

(a). 1 MHz is the frequency of microwave radiation.

(b) 0.167 ms is required by the microwave to travel between two mountains.

Answer:

Explanation:

a. Frequency is the measure of number of times a same thing will be repeated in a given time interval for a given time. And wavelength is the measure of distance between two successive crests or troughs. So wavelength and frequency are inversely proportional to each other. And velocity of light is the proportionality constant.

So frequency of microwave radiation = Speed of light/Wavelength of radiation

Frequency = $\frac{3*10^{8} }{3*10^{-2} }$

Frequency = $10^{8+2} = 10^{10}=10 GHz$

So 10 GHz is the frequency of microwave radiation.

b). As microwave is a part of light waves, so it will be experiencing the speed of light.

As the speed is 3* $10^{8}$ m/s and the distance between the two mountains is given as 50 km, then time can be calculated as

Time = Distance/Velocity

Time = $\frac{50*10^{3} m}{3*10^{8} }=16.67*10^{3-8}=16.67*10^{-5}$

So time = 0.167 ms.

Thus, 0.167 ms is required by the microwave to travel between two mountains.

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3 years ago

Which of the following is a subsurface event takes place during the rock cycle

alukav5142 [94]

Answer:

The answer is A. Cementing...

Explanation:

hope this helps

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3 years ago

Read 2 more answers

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

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3 years ago