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3 years ago

Are diatomic compounds ionic or covalent?

Physics

1 answer:

umka21 [38]3 years ago

8 0

OK so a diatomic compound is covalent because they have an equal amount of <span>Electronegativity </span>

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4. Compare how thermal and electrical energy transfer between objects and give an example of how you can

Leokris [45]

Answer:

4) we take a can of cold soda there is a transfer from hand to can. a current passes through a light bulb, it becomes incandescent and gives ligh

5) the law of energy conservation states that energy is neither created nor destroyed, therefore what happens is that you use energy and transform it into another type of energy,

Explanation:

4) This exercise is asked to give some examples of thermal and electrical energy transfer

the transfer of thermal energy occurs when we grasp a hot or cold body due to the difference in temperature between the body and us there is an exchange of heat, for example when drinking a cup of hot coffee there is a transfer of energy from the cup to the hand .

If we take a can of cold soda there is a transfer from hand to can.

Electrical transfer occurs, for example, when a current passes through a light bulb, it becomes incandescent and gives light.

In all modern electrical appliances there is transformation of electrical energy

5) the law of energy conservation states that energy is neither created nor destroyed, therefore what happens is that you use energy and transform it into another type of energy,

For example, when you lift a box, the potential energy of the box increases with height, but the energy of the person decreases because it does work that is negative.

In amusement park rides, the energy accumulated in one part of the game generally with height is transformed into energy of movement in another part of the game, but the total energy remains constant.

5 0

3 years ago

The multiple reflection of a single sound wave is a/an

ANTONII [103]

A single reflection, like shouting at the side of a mountain and hearing
your voice come back to you, is an 'echo'.

Multiple reflection, like clapping your hands once inside a large room,
is 'reverberation'.

8 0

3 years ago

STATE THE HOOKE'S LAW

victus00 [196]

The Hooke's law is a principal of physics that states that the force needed to extend or compress a spring by some distance x scales linearly with respect to that distance.

6 0

4 years ago

Read 2 more answers

While traveling along a highway a driver slows from 24 m/sec to 15 m/sec in 12 seconds. What is the

Vladimir [108]

24-15=9 m/s slower in 12 seconds. So 9/12 m/s² slower. Therefore the acceleration is -0,75 m/s²

4 0

3 years ago

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:

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

6 0

3 years ago