Ask question

Ask question

All categories

2 years ago

6

The surface of the sun has a temperature of about 5,800 Kelvin. What is the average kinetic energy of particles on the surface o

f the sun? Please show your work

1 answer:

LenaWriter [7]2 years ago

5 0

Answer:

273.15

Explanation:

So that's three over two times 1.38 times ten to the minus twenty-three joules per Kelvin, times 5500 degrees Celsius, the surface of the sun converted into Kelvin by adding 273.15. This works out to 1.20 times ten to the minus nineteen joules. So that's the average kinetic energy of hydrogen atoms.

You might be interested in

For a wave, the _____ the amplitude, the _____ energy the wave carries.

Rudik [331]

Answer:

<h2>HIGHER & MORE OR LARGER OR MORE </h2>

HENCE, THE ANSWER IS A. :)

Explanation:

<em><u>#</u></em><em><u>CARRYONLEARNING</u></em>

<em><u>BRAINLIEST</u></em><em><u> </u></em><em><u> </u></em><em><u>PLEASE</u></em><em><u> </u></em><em><u>I </u></em><em><u>REALLY </u></em><em><u>NEED</u></em><em><u> </u></em><em><u>IT</u></em>

3 0

2 years ago

Read 2 more answers

When an oxygen atom forms an ion, it gains two electrons

kirill [66]

Note the atom of the Oxygen is electrically neutral, meaning it has equal numbers of electrons and protons.

So if it gains 2 electrons, it would have excess of 2 electrons, hence its charge would be -2.

Option B.

5 0

2 years ago

Explain how wind erosion changes land forms

valkas [14]

Answer:

the wind carries abrasive materials

Explanation:

such as sand and salt over time theses small particles slowly strip way at the land form sculpting it by eroding the softer layers first

8 0

2 years ago

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C

miskamm [114]

<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

$\boxed{q_{B}=+2q}$

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential $V$ , which is <em>the energy per unit charge, </em>so the potential $V$ at any point in an electric field with a test charge $q_{0}$ at that point is:

$V=\frac{U}{q_{0}}$

The potential $V$ due to a single point charge q is:

$V=k\frac{q}{r}$

Where $k$ is an electric constant, $q$ is value of point charge and $r$ is the distance from point charge to where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius $r$ . Before the sphere A and B touches we have:

$V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r$

When they touches each other the potential is the same, so:

$V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}$

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

$q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}$

Therefore:

$(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}$

So after A and B touch and are separated the charge on sphere B is:

$\boxed{q_{B}=+2q}$

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

$\boxed{q_{C}=+1.5q}$

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

$q_{A}=q_{B}=+2q$

Second: C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here $q_{A}=+2q$ and C carries no net charge or $q_{C}=0$ . Also, $r_{A}=r_{C}=r$

$V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

Applying the same concept as the previous problem when sphere touches we have:

$k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}$

For the principle of conservation of charge:

$q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q$

Finally, from the third step:

Here $q_{B}=+2q \ and \ q_{C}=+q$ . Also, $r_{B}=r_{C}=r$

$V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

When sphere touches we have:

$k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}$

For the principle of conservation of charge:

$q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q$

So the charge that ends up on sphere C is:

$q_{C}=+1.5q$

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

$+4q$

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

$q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0$

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

$q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q$

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

$+4q$

8 0

3 years ago

An 85-kg refrigerator is located on the 70th floor of a skyscraper (4km above the ground ) what is the potential energy

Airida [17]

Potential energy can be calculated using the following rule:
potential energy = mgh where:
m is the mass = 85 kg
g is the acceleration due to gravity = 9.8 m/sec^2
h is the height = 4 km = 4000 meters

Substitute in the above equation to get the potential energy as follows:
Potential energy = 85*9.8*4000 = 3332000 joules

7 0

3 years ago