The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.
<h3>What is a frequency?</h3>
The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.
Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,
Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.
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Answer:
The total charge Q of the sphere is 
.
Explanation:
Given that,
Radius = 5 cm
Charge density 
We need to calculate the total charge Q of the sphere
Using formula of charge
Where, 
= charge density
V = volume
Put the value into the formula
Put the value into the formula
Hence, The total charge Q of the sphere is .
Answer:
Explanation:
If Bradley examination was done and interpreted in the same facility, the radiologist code is used example- procedure code 72100- Radiologic examination, spine, lumbosacral, 2 or 3 views is reported.
if the X-ray was taken by Dr X but Dr X does not read or interpret the image but forward it to the radiologist for initial report, then a 26- modifier is used. E.g A reports by the technologist would be, procedure code 72050-Radiologic examination, spine, cervical, 2 or 3
views or 72050- TC in certain situations and the consulting radiologist would report 72050-26.
if Bradley’s x-ray were sent to an independent radiologist for interpretation, then the procedure code 76140 is used in reporting.
Answer:
your answer will be 320kg that would be the pressure at the bottom surface of the block
Answer:
610 meters.
Explanation:
Because Jim released the accelerator, the truck started to slow down, so the friction force will eventually stop the truck.
the kinetic energy of the truck just after Jim released the pedal is:
The work done by the friction force is given by:
