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:
Increasing its charge
Increasing the field strength
Explanation:
For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:
where
q is the charge
v is the velocity
B is the magnetic field
m is the mass
r is the radius of the orbit
The period of the motion is
Re-arranging for r
And substituting into the previous equation
Solving for T,
So we see that the period is:
- proportional to the charge and the magnetic field
- inversely proportional to the mass and the square of the speed
So the following will increase the period of the particle's motion:
Increasing its charge
Increasing the field strength
<span>1/3
The key thing to remember about an elastic collision is that it preserves both momentum and kinetic energy. For this problem I will assume the more massive particle has a mass of 1 and that the initial velocities are 1 and -1. The ratio of the masses will be represented by the less massive particle and will have the value "r"
The equation for kinetic energy is
E = 1/2MV^2.
So the energy for the system prior to collision is
0.5r(-1)^2 + 0.5(1)^2 = 0.5r + 0.5
The energy after the collision is
0.5rv^2
Setting the two equations equal to each other
0.5r + 0.5 = 0.5rv^2
r + 1 = rv^2
(r + 1)/r = v^2
sqrt((r + 1)/r) = v
The momentum prior to collision is
-1r + 1
Momentum after collision is
rv
Setting the equations equal to each other
rv = -1r + 1
rv +1r = 1
r(v+1) = 1
Now we have 2 equations with 2 unknowns.
sqrt((r + 1)/r) = v
r(v+1) = 1
Substitute the value v in the 2nd equation with sqrt((r+1)/r) and solve for r.
r(sqrt((r + 1)/r)+1) = 1
r*sqrt((r + 1)/r) + r = 1
r*sqrt(1+1/r) + r = 1
r*sqrt(1+1/r) = 1 - r
r^2*(1+1/r) = 1 - 2r + r^2
r^2 + r = 1 - 2r + r^2
r = 1 - 2r
3r = 1
r = 1/3
So the less massive particle is 1/3 the mass of the more massive particle.</span>
I don't know what the exact word is, but I do know that the bigger an objects mass is the more it will attract other objects toward it, mainly smaller objects with less mass. it might be gravity or something around those lines....is it a multiple choice question?
KE = (1/2)·(mass)·(speed)²
KE = (1/2)·(50 kg)·(18 m/s)²
KE = (25 kg)·(324 m²/s²)
KE = 8,100 kg-m²/s²
KE = 8,100 Joules
