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3 years ago

Does potential energy increase with temperature?

2 answers:

6 0

-- The potential energy of a 12-lb bowling ball up on the shelf
doesn't have anything to do with the temperature of the ball or
the shelf.

-- The potential energy of a jar full of gas does depend on the
temperature of the gas. The warmer it is, the greater its pressure
is, and the more work it can do if you let it out through a little hole
in the jar. If it gets hot enough, it'll have enough potential energy
to blow the jar to smithereens.

Wewaii [24]3 years ago

3 0

When a solid melts and becomes a liquid, we say it changes phase from a solid to a liquid. In this change, the bonding between the atoms or molecules changes. You have to "break" some bonds to go from a solid to a liquid. This requires energy. The liquid is a "higher" potential energy state than the solid, even at the same temperature. (It is slightly more complicated than this, but this is good enough for this class.) To convert 1kg of solid water at 0oC (273K) to liquid water at 0oC (273K) requires about 330,000J of energy. Note that the temperature of the liquid is the same as the solid’s, i.e. you added heat without changing the temperature, instead the phase changed. The heat added went into "breaking" bonds and increasing its potential energy, not into increasing the average translational KE of the molecules. (It is slightly more complicated than this, but this explanation is good enough for this class.) If you go the other way, and convert 1kg of liquid water at 0oC (273K) to ice at 0oC (273K) releases 330,000J of heat. This heat comes from the energy given off when bonds form, i.e. it goes to a state of lower potential energy. The same type of thing occurs when a liquid changes to a gas. Then more bonds are broken as the molecules move apart, and it requires energy to break the bonds and move to a higher potential energy. To convert 1kg of liquid water at 100oC (373K) to 1kg of water vapor at 100oC (373K) requires 2,260,000J (almost 2.3 million Joules) of energy. That is, the water absorbs energy to change from a liquid to a gas.. If 1kg of water vapor conde

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Write a letter to a Friend congratulating her / him for her excellent result in annual examination. class 5 the

gulaghasi [49]

Answer:

ok will do that

Explanation:

7 0

3 years ago

An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. Aft

sergeinik [125]

Answer:

a) 80 V

b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

Explanation:

Given :

We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( $K_{A}$ = 0) to the point B at distance l = 0.500m where its kinetic energy is ( $K_{B}$ = 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = + 30.0 v.

Required :

(a) We are asked to find the electric potential VB

(b) We want to determine the magnitude and the direction of the electric field E.

Solution

(a) We are given the values for VA, $K_{B}$ and q so we want to find a relationship between these three parameters and VB to get the value of VB.

As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:

$K_{A} +U_{A} =K_{B} +U_{B}$ .........................................(1)

Where $K_{A}$ = 0 and the potential energy U of the charge is given by U = q V

where V is the electric potential. So, equation (1) will be in the form :

0+qVA= $K_{B}$ +qVB (Divide by q)

VA= $K_{B}$ /q + VB (solve for VB)

VB=VA- $K_{B}$ /q .......................................(2)

We get the relation between VB, VA and $K_{B}$ , now we can plug our values for VA, $K_{B}$ and q into equation (2) to get VB

VB=VA- $K_{B}$ /q

=30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)

=80 V

(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points

VA-VB = $\int\limits^1_0 {E} \, dl$ ...................................(a)

$=E\int\limits^1_0 {} \, dl$

VA-VB=E $l$ (solve for E)

E= VA-VB/ $l$ ..................................(3)

Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E

E= VA-VB/ $l$

=-100 N/C

The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

5 0

3 years ago

What makes a good scientific question? (1 point)

liraira [26]

C. It is answered by observation and evidence.

Good scientific explanations are defined, measurable and controllable. They can be answered by an experiment.

7 0

3 years ago

All please..............

rodikova [14]

4 water has a low heat capacity and a high vaporization temperate and coasts have low temps

3 0

3 years ago

Describes the relationship between the spectrum of one atom and the spectrum of another?

krok68 [10]

The relationship between the number of visible spectral lines are identical for atoms .However they have unique wavelengths.

Option B

Explanation:

A spectrum is a range of frequencies or a range of wavelengths. The photon energy of the emitted photon is equal to the difference between two states. For every atom there are quite many electron transitions and each has a energy difference.

This difference in wavelength causes spectrum .As each element emission spectrum is unique because each atom has different energy and causes uniqueness in the emission spectrum . Hence, due to the difference in energy it emits different wavelengths.

6 0

3 years ago