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2 years ago

PLZZ I WILL DO ANY THING

Physics

1 answer:

zhuklara [117]2 years ago

3 0

Answer: Universe.

Explanation: In addition to the solar system, the heliocentric model stated that the sun was the center of the UNIVERSE.

Every plant orbits around the sun, which makes our universe whole and bright.

Hope this helps you out! ☺

-Karleif-

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9) Of all the types of light the Sun gives off, it emits the greatest amount of light at visible wavelengths of light. If the Su

erastova [34]

Answer:

* most of the emission would be in the infrared part, the visible radiation would be very small.

*total intensity of the semition decreases that the intensity depends on the fourth power of the temperature

Explanation:

The radiation emitted by the Sun is approximately the radiation of a black body, if the Sun were to cool, the maximum emission wavelength changes

λ T = 2,898 10⁻³

λ = 2,898 10⁻³ / T

if the temperature decreases the maximum wavelength the greater values are moved, that is to say towards the infrared. Therefore the emission curve also moves, in this case most of the emission would be in the infrared part, the visible radiation would be very small.

Furthermore, the total intensity of the semition decreases that the intensity depends on the fourth power of the temperature according to Stefan's law

P = σ A eT⁴

7 0

3 years ago

The following are secondary sources of energy except a.coal b crude oil c nuclear d wind e wood

monitta

Answer:

d) Wind

Explanation:

Secondary energy is energy produced by converting energy available in its natural state in the environment. Hence Wind is a primary source not a secondary source

8 0

2 years ago

A bike rider pedals with constant acceleration to reach a velocity of 7.5 (vf)m/s over a time of 4.5 s(t). during the period of

BigorU [14]

<span>The initial velocity of the bike was 1.67 (vf)m/s. This is found by evaluating 7.5/4.5 which yields the velocity per unit of time which is equivalent to initial velocity.</span>

6 0

3 years ago

What is meant by angular frequency?

andrew11 [14]

measures angular displacement per unit time. Its units are therefore degrees (or radians) per second.

6 0

3 years ago

What is the shortest-wavelength x-ray radiation in m that can be generated in an x-ray tube with an applied voltage of 93.3 kV?

VikaD [51]

(a) $1.33\cdot 10^{-11} m$

The x-rays in the tube are emitted as a result of the collisions of electrons (accelerated through the potential difference applied) on the metal target. Therefore, all the energy of the accelerated electron is converted into energy of the emitted photon:

$e \Delta V = \frac{hc}{\lambda}$

where the term on the left is the electric potential energy given by the electron, and the term on the right is the energy of the emitted photon, and where:

$e=1.6\cdot 10^{-19}C$ is the electron's charge

$\Delta V = 93.3 kV = 93300 V$ is the potential difference

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

$c=3.00\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength of the emitted photon

Solving the formula for $\lambda$ , we find:

$\lambda=\frac{hc}{e\Delta V}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{(1.6\cdot 10^{-19})(93300)}=1.33\cdot 10^{-11} m$

(b) 93300 eV (93.3 keV)

The energy of the emitted photon is given by:

$E=\frac{hc}{\lambda}$

where

h is Planck constant

c is the speed of light

$\lambda=1.33\cdot 10^{-11} m$ is the wavelength of the photon, calculated previously

Substituting,

$E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.33\cdot 10^{-11}}=1.50\cdot 10^{-14} J$

Now if we want to convert into electronvolts, we have to divide by the charge of the electron:

$E=\frac{1.50\cdot 10^{-14} J}{1.6\cdot 10^{-19} J/eV}=93300 eV$

(c) The following statements are correct:

The maximum photon energy is just the applied voltage times the electron charge. (1)

The value of the voltage in volts equals the value of the maximum photon energy in electron volts.

In fact, we see that statement (1) corresponds to the equation that we wrote in part (a):

$e \Delta V = \frac{hc}{\lambda}$

While statement (2) is also true, since in part (b) we found that the photon energy is 93.3 keV, while the voltage was 93.3 kV.

3 0

3 years ago