Answer:
Wavelength, frequency and the photon energy changes as the one goes across the ranges of the electro-magnetic radiations.
Explanation:
Electro-magnetic radiations may be defined as the form of energy that is radiated or given by the electro-magnetic radiations. The visible light that we can see is the one of the electro-magnetic radiations. Other forms are the radio waves, gamma waves, UV rays, infrared radiations, etc.
The wavelength of the radiations decreases as we go from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
The frequency of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
The photon energy of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
Answer:
change in entropy is 3.3034 ×
Explanation:
give data
thermal energy Q = 155 J
temperature T = 340 K
to find out
change in entropy
solution
we know change in entropy formula that is
change in entropy = Q / ( K×T ) ..............1
here K is boltzmann constant that is 1.38 ×
kg-m²/s²
put these value in equation 1 we get
change in entropy = Q / ( K × T )
change in entropy = 155 / ( 1.38 ×
× 340 )
change in entropy = 3.3034 ×
so change in entropy is 3.3034 ×
A 2 µC charge q1 and a 2 µC charge q2 are 0.3 m from the x-axis. A 4 µC charge q3 is 0.4 m from the y-axis. The distances d13 and d23 are 0.5 m. Find the magnitude and direction of the resulting vector R. Round your answer to the nearest tenth.
The magnitude is 0.5 N.
The direction is °.0
ON EDGENUTITY
Answer:
Explanation:
The work W required for the transfer is equal to the energy difference
![W = \Delta E = q \ \Delta V](https://tex.z-dn.net/?f=W%20%3D%20%5CDelta%20E%20%3D%20q%20%5C%20%5CDelta%20V)
where q is the charge we want to transfer and ΔV is the potential difference.
So, the work needed will be
![W = 9,000 \ C * 220 V](https://tex.z-dn.net/?f=W%20%3D%209%2C000%20%5C%20C%20%2A%20220%20V)
![W = 1,980,000 \ J](https://tex.z-dn.net/?f=W%20%3D%201%2C980%2C000%20%5C%20J)
Now, the average power is:
![< power > = \frac{work}{time}](https://tex.z-dn.net/?f=%3C%20power%20%3E%20%3D%20%5Cfrac%7Bwork%7D%7Btime%7D)
As the Watt is Joule/second, we need the 45 min multiplied by 60:
![time = 45 \ min * 60 \frac{s}{min}](https://tex.z-dn.net/?f=time%20%3D%2045%20%5C%20min%20%2A%2060%20%5Cfrac%7Bs%7D%7Bmin%7D)
![time = 2,700 \ s](https://tex.z-dn.net/?f=time%20%3D%202%2C700%20%5C%20s)
Taking all this together:
![< power > = \frac{1,980,000 \ J}{2,700 \ s}](https://tex.z-dn.net/?f=%3C%20power%20%3E%20%3D%20%5Cfrac%7B1%2C980%2C000%20%5C%20J%7D%7B2%2C700%20%5C%20s%7D)
![< power > = \frac{1,980,000 \ J}{2,700 \ s}](https://tex.z-dn.net/?f=%3C%20power%20%3E%20%3D%20%5Cfrac%7B1%2C980%2C000%20%5C%20J%7D%7B2%2C700%20%5C%20s%7D)
![< power > = 733.333 \ W](https://tex.z-dn.net/?f=%3C%20power%20%3E%20%3D%20733.333%20%5C%20W)
or
![< power > = 0.733 \ kW](https://tex.z-dn.net/?f=%3C%20power%20%3E%20%3D%200.733%20%5C%20kW)
Answer:
Ok I’m not pretty sure but I think it’s C but look for more answers first before you put C because I don’t want you to get it wrong. :)
Explanation: