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What is the equation for potential energy

meriva

PE = mgh
where
m = mass
g = acceleration due to gravity
h = height

3 0

2 years ago

Which statement is true about effective nuclear charge?a. Effective nuclear charge decreases as we move to the right across a ro

MissTica

Answer:

Option b. Effective nuclear charge increases as we move to the right across a row in the periodic table

Explanation:

The effective nuclear charge is a measure of how strong the protons in the nucleus of an atom attract the outermost electrons of such atom.

The effective nuclear charge is the net positive charge experienced by valence electrons and is calculated (as an approximation) by the equation: Zeff = Z – S, where Z is the atomic number and S is the number of shielding electrons.

The shielding electrons are those electrons in between the interesting electrons and the nucleus of the atom.

Since the shielding electrons are closer to the nucleus, they repel the outermost electrons and so cancel some of the attraction exerted by the positive charge of the nucleus, meaning that the outermost electrons feel less the efect of attraction of the protons. That is why in the equation of Zeff, the shielding electrons (S) subtract the total from the atomic number Z.

The effective nuclear charge, then, is responsible for some properties and trends in the periodic table. Here, you can see how this explains the trend of the atomic radius (size of the atom) accross a row in the periodic table.

As the effective nuclear charge is larger, in a same row of the periodic table, the shielding effect is lower, the outermost electrons are more strongly attracted by the nucleus, and the size of the atoms decrease. That is why as we move to the right in the periodic table, the size of the atoms decrease.

3 0

3 years ago

Which of the following is a characteristic of inner planets? (2 points)

Svetradugi [14.3K]

Few moons
I just took the test and got it right

7 0

3 years ago

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A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

a)

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

b)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

c)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

where:

$q=+4.9\cdot 10^{-6}C$ is the charge

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

$F_{B_y}=qvB_y$

And by substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0

3 years ago

ATP molecules store cellular energy. When phosphate groups are removed and molecules of ADP are formed, energy is released. This

slava [35]

Adenosine triphosphate (ATP) consists of an adenosine molecule bonded to three phophate groups in a row. In a process called cellular respiration, chemical energy in food is converted into chemical energy that the cell can use, and stores it in molecules of ATP. This occurs when a molecule of adenosine diphosphate (ADP) uses the energy released during cellular respiration to bond with a third phosphate group, becoming a molecule of ATP. So the energy from cellular respiration is stored in the bond between the 2nd and 3rd phosphate groups of ATP. When the cell needs energy to do work, ATP loses its 3rd phosphate group, releasing energy stored in the bond that the cell can use to do work. Now its back to being ADP and is ready to store the energy from respiration by bonding with a 3rd phosphate group. ADP and ATP constantly convert back and forth in this manner.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

3 0

3 years ago

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