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2 years ago

Which particles contribute to the net charge and how does each change the net charge?

Physics

1 answer:

Licemer1 [7]2 years ago

3 0

The two subatomic particles that contribute to the net charge of an ion are electrons and protons.

Atom is the smallest possible amount of matter which still retains its identity as a chemical element, now known to consist of a nucleus surrounded by electrons.

The atom is made up of three components called subatomic particles as follows;

Protons
Electrons
Neutrons

The proton is the positively charged subatomic particle forming part of the nucleus of an atomwhile the electron is the subatomic particle having a negative charge and orbiting the nucleus.

This suggests that the two subatomic particles that contribute to the net charge of an ion are electrons and protons. That is;

Net charge = protons - electrons

Learn more about subatomic particles at:brainly.com/question/13303285

#SPJ1

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The elastic portion of the downward-sloping straight-line demand curve lies:_______

zepelin [54]

Answer:

c. above the point of unit elasticity.

Explanation:

The elastic portion of the downward-sloping straight-line demand curve lies above the point of unit elasticity. Supply and demand are fundamental concept in economics. The demand curve shows how much of a good people will want at a different prices. The demands curves illustrates the intuition why people purchase a good for a lower price. For the demand curve, the price is always shown on the vertical axis and the demand curve is shown on the horizontal axis. Thus , the quantity demanded increases as the price gets lower. However, the price elasticity of the demand curve varies along the demand curve. This is because there is a key distinction between the gradient and the elasticity. The gradient which is the slope of the line is always the same in the demand curve but elasticity of the demand changes in the percentage of the quantity demand. Therefore, elasticity will vary along the downward-sloping straight - line demand curve. So, in a downward-sloping straight-line demand curve, the elastic portion is usually above the point of unit elasticity

7 0

3 years ago

A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice

Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of $\sqrt{3} v_{e}$

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , $v_{i}=2 v_{e}$ . Here the radius is R, so find the escape velocity as follows,

$\frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0$

$\frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}$

$v_{e}^{2}=\frac{2 G M}{R}$

$v_{e}=\sqrt{\frac{2 G M}{R}}$

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as, $U_{i}=-\frac{G M m}{R}$

R tend to infinity when far away from the planet, so $v_{f}=0$

Then, kinetic energy at initial would be,

$k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}$

Similarly, kinetic energy at final would be,

$k_{f}=\frac{1}{2} m v_{f}^{2}$

Here, $v_{f}=\text { final velocity }$

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

$U_{i}+k_{i}=k_{f}+v_{f}$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}$

'm' and $\frac{1}{2}$ as common on both sides, so gets cancelled, we get as

$4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}$

We know, $v_{e}=\sqrt{\frac{2 G M}{R}}$ , it can be wriiten as $\left(v_{e}\right)^{2}=\frac{2 G M}{R}$ , we get

$4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}$

$v_{f}^{2}=3\left(v_{e}\right)^{2}$

Taking squares out, we get,

$v_{f}=\sqrt{3} v_{e}$

4 0

3 years ago

If you ride your bike 20 miles and it takes you 120 minutes,what is your average speed

Alexus [3.1K]

120 minutes=2 hours
20/2= 10mph

5 0

3 years ago

Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such li

lozanna [386]

Answer: 4.50*10^-6T (0.00000450071T)

Explanation: A current carrying conductor has been knowing to generate a specific amount of magnetic field.

This is given by the Bio-savart law (mathematical).

The Bio-savart law is a mathematical equation that gives the value of strength of the magnetic field created by a current carrying conductor.

B=(Uo* I) /2πr

Where

B= strength of magnetic field

Uo = magnetic permeability in free space = 1.257 *10^-6

r = distance between current carrying conductor and any reference point.

By doing the neccesary algebra, we have

B=(1.257 *10^-6 * 180)/ (2 * 3.142 * 8)

B= 2.2626 *10^-4 / 50.2857

B=4.5 * 10^-6T (0.00000450071T)

6 0

3 years ago

Universal ethical standards also apply to international businesses. <br> a. true <br> b. false

Rzqust [24]

I would pick true...

5 0

4 years ago