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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

here we have

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

$U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})$

$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0

2 years ago

Why do we get dizzy when we spin?

iris [78.8K]

The body senses whether it is upright or lying down or whether it is moving or standing still through the vestibular system, which is in the upper portion of the inner ear.

3 0

3 years ago

In a binary star system, an unseen component is found to have 8 solar masses. It would be visible if the system were a normal st

maksim [4K]

Answer:

Black Hole

Explanation:

A black hole is a very dense and massive stellar object, which has a field of gravity so large that not even light can escape it.

Since it does not emit light, we cannot see them directly, hence the name of black hole.

So in this case, if the object has a mass of 8 solar masses that is enough to form a black hole, and also cannot be seen, all of this indicates that the object we are talking about is a black hole.

It should be mentioned that although these objects do not emit light, because it cannot escape due to the immense force of gravity, black holes can be detected by a type of radiation emitted on their event horizon due to quantum effects called Hawking radiation .

5 0

3 years ago

In Ch. 1.6, the authors point out that interstellar space is not actually as empty as it seems. There is actually a lot of diffu

wel

Answer:

very small solid particles called interstellar dust.

Explanation:

In the space between the stars there is gas and dust, which represent at least 20% of the mass of our galaxy. In the Milky Way it is considered that there is a gas density of approximately 0.2 to 0.5 atoms / cm3 in the surroundings of the Sun; with respect to the dust an average of 1 g / cm3 is estimated.

Gas is about atoms and molecules, mainly hydrogen; In order of abundance, helium, carbon, oxygen, nitrogen and iron follow. On the other hand, the dust is tiny particles, generally smaller than 10 microns; the dust does not shine and therefore it is only distinguished when it is projected on bright regions (nebulae or clusters).

Interstellar matter is mainly concentrated towards the plane of the galaxy, in the strip corresponding to the Milky Way; there you can see bright nebulas of diffuse character called nebulas. These nebulae are classified according to three types: (a) bright or emission nebulae, (b) reflection nebulae and (c) planetary nebulae.

Hydrogen appears both ionized and neutral; The bright nebulae are composed of ionized hydrogen and other ionized elements. Non-ionized (neutral) hydrogen is found in the spiral arms of the Milky Way and can be detected through radio waves.

6 0

3 years ago

How much must a woman weigh ( force) if the pressure she exerts while standing on one foot has an area of 0.6m squared exerts a

lions [1.4K]

Answer:

W = 9.6 N

Explanation:

Given that,

Area on 1 foot, A = 0.6 m²

Pressure, P = 16 Pa

The pressure is given by force acting per unit area. So,

$P=\dfrac{F}{A}\\\\P=\dfrac{W}{A}\\\\W=16\times 0.6\\\\W=9.6\ N$

So, the required weight is 9.6 N.

4 0

3 years ago