Answer: 8.1 x 10^24
Explanation:
I(t) = (0.6 A) e^(-t/6 hr)
I'll leave out units for neatness: I(t) = 0.6e^(-t/6)
If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).
For neatness let k = 1/(6x3600) = 4.63x10^-5, then:
I(t) = 0.6e^(-kt)
Providing t is in seconds, total charge Q in coulombs is
Q= ∫ I(t).dt evaluated from t=0 to t=∞.
Q = ∫(0.6e^(-kt)
= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.
= -(0.6/k)[e^-∞ - e^-0]
= -0.6/k[0 - 1]
= 0.6/k
= 0.6/(4.63x10^-5)
= 12958 C
Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.
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.
A complex entity involving the Earth's biosphere, atmosphere, oceans, and soil; the totality constituting a feedback or cybernetic system which seeks an optimal physical and chemical environment for life on this planet
We need the frequency of the photon, it is v = c/ λ
Where c is 3 x 10^8 ms^-1 and λ
is the wave length
We also need the expression of
connecting frequency to energy of photon
which is E = hv where h is Planck’s
constant
Combining the two equations
will give us:
E = h x c/λ
Inserting the values, we will
have:
E = 6.626 x 10^-34 x 3 x 10^8 /
0.126
E = 1.578 x 10^ -24 J
