Question 1

Helga [31]

Nuclear energy was not developed. It has existed for as long as time has existed, that is, since the big bang.

The thing that was developed was humans' ability to USE nuclear energy, to do what we want it to do, when we want it to do it.

The reason this was first developed was to bomb the holy beans out of Japan, in order to win World War II.

Today (2020), nine of the world's nations are known to have 14,285 nuclear bombs in storage, for the same general purpose. Seven of these nations are storing 1,170 of these bombs (about 8 percent), and the USA and Russia have all the rest ... 13,035 nuclear bombs.

All nine of these nations promise that they have no plan to use their bombs, they don't want to use them, it would be wrong and terrible to use them, and they will never be the first to use them, but they need to modernize their bombs so that theirs are better than anybody else's bombs, and they need to keep their bombs for as long as anybody else has any, and then maybe a little longer, just in case.

In the years after the ability to bomb the holy beans out of other people was developed, and enough equipment was built to do it 14 thousand times, the ability to use nuclear energy for other purposes was also developed. It's used now to generate electrical energy, and to do several jobs in Medical science.

7 0

3 years ago

Precision measurements of the acceleration due to gravity show that the acceleration is slightly different in different location

DochEvi [55]

Gravity largely depends on the comparison of two objects; it's why you have the equation F= (GMm)/r^2. On Earth, you have different altitudes that, with the formula, will give different results for gravity because the radius is different everywhere. This difference on calculations, however, are seen to be miniscule. We know gravity as 9.81 m/s^2 but it might be different by thousandths or hundreds of thousandths of a decimal.

6 0

3 years ago

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The distance between a charge and the source of an electric field changes from 3 mm to 6 mm. as a result of the change, the elec

STatiana [176]

The electric potential energy of the charge is reduced because it decreases with increase in the distance between charges.

<h3>What is electric potential energy?</h3>

Electric potential energy can be defined as the energy needed to move a charge against an electric field.

It is calculated using the formula;

U = Kq1 q2 ÷ r

Where Q = electric potential energy

k = Coulombs constant

q1 and q2 = charges

r = distance of separation

Electric potential energy is inversely proportional to the distance of separation of the charges.

If the distance of the charges changes from 3mm to 6mm, then the electric potential energy of the charges is reduced because it decreases with increase in the distance of the charges.

Therefore, the electric potential energy of the charge is reduced because it decreases with increase in the distance between charges.

Learn more about electric potential energy here:

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4 0

1 year ago

Like charges will exert a force of

natulia [17]

Answer:

D- Repulsion

Explanation:

A positively charged object will exert a repulsive force upon a second positively charged object.

7 0

3 years ago

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What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec

jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

$E = \sigma T^{4}$ (1)

Where $\sigma$ is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

$\lambda max = \frac{2.898x10^{-3} m. K}{T}$ (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

$E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}$ (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

$T = \frac{2.898x10^{-3} m. K}{\lambda max}$ (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), $\lambda max$ (450 nm) will be express in meters:

$450 nm . \frac{1m}{1x10^{9} nm}$ ⇒ $4.5x10^{-7}m$

$T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}$

$T = 6440 K$

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

$700 nm . \frac{1m}{1x10^{9} nm}$ ⇒ $7x10^{-7}m$

$T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}$

$T = 4140 K$

Comparison:

$\frac{6440 K}{4140 K} = 1.55$

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0

3 years ago