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3 years ago

In an electron dot diagram, each dot represents a(an ____________________.

1 answer:

8 0

Each dot represents a valence electron. Valence electrons are the electron on the outer electron shell of an atom.<span>
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Please in the name of god help me guys.

zepelin [54]

1. D: the ability to cause a change or do work

The others dont make sense for energy.

2. C: Lamp

A lamp uses the electric current to light up a room, or area.

3. B: foil will bend away

The same charges repel each other that is why a magnet is repelled by the same pole of another magnet.

4. A: Sound

You can’t hear light or any of the other options.

5. A: gravity

Gravity is a force of nature that pulls on everything.

6. C: static electricity

The opposite charges stick to each other, a battery again is an example of this.

7. B: a copper nail

Metal is one of the best electric conductors.

8. C: closed circuit

This is because if a circuit is closed the electricity flows through perfectly but for example an open one would not work or it would leak energy.

Hope this helps

3 0

3 years ago

Read 2 more answers

If a body of 2kg mass is at a distance of 7200km from the center of the earth .What

wlad13 [49]

None of the choices is correct.

== The Earth's radius is 6,371 km. That's how far you are from the center of the Earth when you're on the ground.

== The acceleration of gravity on the ground is 9.8 m/s² .

== The acceleration of gravity varies inversely as the square of the distance from the center of the Earth. So, at 7,200 km from the center, it's

(9.8 m/s²) · (6371km/7200km)² .

Stuffing this through your cakulator, you'll get <em>7.67 m/s²</em> .

3 0

3 years ago

A charge of -8.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.05 cm.

gavmur [86]

Answer:

(a) $E = -1.02 \times 10^5~N/C$

(b) $E = -9.7 \times 10^4~N/C$

Explanation:

(a) The electric field for a point charge is given by the following formula:

$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r$

Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(\sqrt{z^2 + r^2})^2}$

Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.

In cylindrical coordinates, da = r dr dθ. So,

$\frac{Q}{\pi R^2} = \frac{dQ}{da}\\\frac{Q}{\pi R^2} = \frac{dQ}{rdrd\theta}\\dQ = \frac{Qrdrd\theta}{\pi R^2}$

Hence, dE is now:

$dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}$

The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.

It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.

Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,

$\sin(\alpha) = \frac{z}{\sqrt{z^2 + r^2}}$

Therefore, vertical component of dE now becomes

$dE_z = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}\frac{z}{\sqrt{z^2+r^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 \int\limits^R_0 {\frac{rdrd\theta}{(z^2+r^2)^{3/2}}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2} 2\pi(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})$

Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:

$E_z = \frac{1}{2\epsilon_0}\frac{Qz}{\pi R^2}(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}) = -1.02 \times 10^5~N/C$

(b) We will have a similar approach, but a simpler integral.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{z^2 + R^2}\\\frac{Q}{2\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{2\pi}\\dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qzd\theta}{2\pi(z^2 + R^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}2\pi$

$E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}} = -9.07\times 10^4~N/C$

Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.

3 0

3 years ago

A spring whose stiffness is 1140 n/m has a relaxed length of 0.51 m. if the length of the spring changes from 0.26 m to 0.79 m,

Ugo [173]

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.
Read more on Brainly.com - brainly.com/question/1581851#readmore

7 0

4 years ago

A ladder is leaning against a building so that the distance from the ground to the top of the ladder is 8 feet less than the len

SVETLANKA909090 [29]

This is the concept of application of Pythagorean theorem and algebra;
Suppose the the length of the ladder is x ft, the distance from the ground to the top of the ladder will be (x-8) ft, the distance from the bottom of the ladder to the building is 16ft, thus to find the length of the ladder we proceed as follows;
c^2=a^2+b^2
x^2=(x-8)^2+16^2
x^2=x^2-16x+64+256
collecting the like terms we get:
x^2-x^2+16x=320
16x=320
solving for x we get:
x=320/16
x=20 ft
the answer is x=20 ft

7 0

4 years ago