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3 years ago

Using the vocabulary words we have learned so far in this unit, describe what happens when something warms up.

2 answers:

4 0

When heat is added to a substance, the molecules and atoms vibrate faster. As atoms vibrate faster, the space between atoms increase. The motion and spacing of the particles determines the state of matter of the substance. The end result of increased molecular motion is that the object expands and takes up more space.

Lunna [17]3 years ago

4 0

The answer is molecule

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Absorbed energy is said to be quantized. This means:____________ a) Energy is absorbed only by valence electrons. b) Energy caus

wlad13 [49]

Answer:Energy is absorbed in exact amounts that correspond to differences in energy levels.

Explanation:

According to Bohr's theory, an electron moves from a lower to a higher energy level when it absorbed photon of a given energy. The energy of the photon absorbed must be equal to the energy difference between the higher and lower energy level. Thus energy is absorbed only in discreet amounts that correspond to differences in energy levels. Absorption of energy in discreet amounts is what is meant by the term 'quantization'.

4 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

17.

Oliga [24]

I think potential but im not sure

8 0

2 years ago

Just after two identical point charges are released when they are a distance D apart in outer space, they have an acceleration a

Darya [45]

Answer:

d) 4a

Explanation:

r = Distance

Electrostatic force is given by

$F=\dfrac{kq_1q_2}{r^2}$

$ma=\dfrac{kq_1q_2}{r^2}$

It can be seen that the force is inversely proportional to distance

$a\propto \dfrac{1}{r^2}$

If the distance is reduced

$\dfrac{a_1}{a_2}=\dfrac{r_2^2}{r_1^2}\\\Rightarrow \dfrac{a_1}{a_2}=\dfrac{(\dfrac{d}{2})^2}{d^2}\\\Rightarrow \dfrac{a_1}{a_2}=\dfrac{1}{4}\\\Rightarrow a_2=4a_1$

So, the new acceleration will be four times the old acceleration

The answer is d) 4a

7 0

4 years ago

Entific findings are useful

Kaylis [27]

That's true. The give-away is "observation ... in the real world". That's what "field reports" are.

The part I'm not so sure about is those "entific findings".

4 0

3 years ago