If an atom contains 13 protons, then it has (2.4)a.13 electrons. b. 26 electrons. c. 13 neutrons. d.26 neutrons.

What electric field strength would store 12.5 JJ of energy in every 6.00 mm3mm3 of space?

Tcecarenko [31]

To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as

The potential energy, $U = 13.0 J$

The volume, $V = 6.00 mm^3 = 6.00*10^{-9}m^3$

The potential energy per unit volume is defined as the energy density.

$u = \frac{U}{V}$

$u= \frac{(13.0 J)}{(6.00*10^{-9} m^3)}$

$u= 2.167109 J/m^3$

The energy density related with electric field is given by

$u = \frac{1}{2} \epsilon_0 E^2$

Here, the permitivity of the free space is

$\epsilon_0 = 8.85*10^-{12} C^2/N \cdot m^2$

Therefore, rerranging to find the electric field strength we have,

$E = \sqrt{\frac{2u}{\epsilon_0}}$

$E = \sqrt{\frac{2(2.167109)}{8.85*10^{-12}}}$

$E = 2.211010 V/m$

Therefore the electric field is 2.21V/m

8 0

3 years ago

50 points if brainliest !!!!!!!!!!

Natasha_Volkova [10]

Answer:

$weight = mg \\ = 70 \times 10 \\ = 700 \: newtons$

$force = 700 \: newtons$

$force = mass \times acceleration \\ 700 = 70 \times a \\ a = 10 \: {ms}^{ - 2}$

4 0

3 years ago

Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.92×10−11 m around a stationary p

DaniilM [7]

Answer:

2.068 x 10^6 m / s

Explanation:

radius, r = 5.92 x 10^-11 m

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.

centripetal force = $\frac{mv^{2}}{r}$

Electrostatic force = $\frac{kq^{2}}{r^{2}}$

where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2

So, balancing both the forces we get

$\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}$

$v=\sqrt{\frac{kq^{2}}{mr}}$

$v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}$

v = 2.068 x 10^6 m / s

Thus, the speed of the electron is give by 2.068 x 10^6 m / s.

6 0

3 years ago

A car of mass 1535 kg collides head-on with a parked truck of mass 2000 kg. Spring mounted bumpers ensure that the collision is

dexar [7]

A car of mass 1535 kg collides head-on with a parked truck of mass 2000 kg. Spring mounted bumpers ensure that the collision is essentially elastic. If the velocity of the truck is 17 km/h (in the same direction as the car's initial velocity) after the collision, what was the initial speed of the car <u>20kmh</u>

<h3>What is collision ?</h3>

A collision in physics is any situation in which two or more bodies quickly exert forces on one another. Despite the fact that the most common usage of the word "collision" refers to situations in which two or more objects clash violently, the scientific usage of the word makes no such assumptions.

The following are a few instances of physical encounters that scientists might classify as collisions:

Legs of an insect are said to collide with a leaf when it falls on one.
Every contact of a cat's paws with the ground while it strides across a lawn is seen as a collision, as is every brush of its fur with a blade of grass.

To learn more about collision from the given link:

brainly.com/question/27736776

#SPJ4

7 0

1 year ago

10. Cumulus and stratus clouds belong to which cloud group? <br><br>

EastWind [94]

Cumulus belongs to vertical clouds and status to low

5 0

4 years ago