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:
g_x = 3.0 m / s^2
Explanation:
Given:
- Change in length of spring [email protected] = 22.6 cm
- Time taken for 11 oscillations t = 19.0 s
Find:
- The value of gravitational free fall g_x at plant X:
Solution:
- We will assume a simple harmonic motion of the mass for which Time is:
T = 2*pi*sqrt(k / m ) ...... 1
- Sum of forces in vertical direction @equilibrium is zero:
F_net = k*x - m*g_x = 0
(k / m) = (g_x / x) .... 2
- substitute Eq 2 into Eq 1:
2*pi / T = sqrt ( g_x / x )
g_x = (2*pi / T )^2 * x
- Evaluate g_x:
g_x = (2*pi / (19 / 11) )^2 * 0.226
g_x = 3.0 m / s^2
Explanation:
The time period rotation of the bob is t and the tension in the thread is T. ... A simple pendulum of length. ... Balancing the forces in Horizontal and vertical direction: ... Circular motion 2.
1 answer
Answer:
Fractional error = 0.17
Percent error = 17%
F = 112 ± 19 N
Explanation:
Plug in the values to find the force:
F = (3.5 kg) (20 m/s)² / (12.5 m) = 112 N
Find the fractional error:
ΔF/F = Δm/m + 2Δv/v + Δr/r
ΔF/F = 0.1/3.5 + 2(1/20) + 0.5/12.5
ΔF/F = 0.17
Multiply by 100% to find the percent error:
ΔF/F × 100% = 17%
Solve for the absolute error:
ΔF = 0.17 × 112 N = 19 N
Therefore, the force is:
F = 112 ± 19 N
Answer:
130 km at 35.38 degrees north of east
Explanation:
Suppose the HQ is at the origin (x = 0, y = 0)
So the coordinates of the helicopter after the 1st flight is
After the 2nd flight its coordinate would be:
So in order to fly back to its HQ it must fly a distance and direction of
north of east
