Answer:
Mammography is the process in which low energy radiations are used to diagnose and screening. The purpose of this process is the early detection of the breast cancer. These low energy radiations may have some risks like damaging and burning of cells.
In the current scenario, woman is apprehensive because she has read about the risks of using ionizing radiations. The radiographer should tell her the benefits of the mammography will outweigh its potential consequences. Screening, for instance, will let her know if she is suffering from breast cancer. Cancer is very dangerous disease as compare to very small burning.
In this way radiographer should handle the situation.
Answer:
50N
Explanation:
Force (N) = mass (kg) × acceleration (m/s²)
0.25kg times 200m/s² = 50N
all the allials must be aligned in the same direction
magnets are affected by heat, drops, and improper storage
Answer:
b) 472HZ, 408HZ
Explanation:
To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:

fo: frequency of the source = 440Hz
vs: speed of sound = 343m/s
vo: speed of the observer = 0m/s (at rest)
v: sped of the train
f: frequency perceived when the train leaves us.
f': frequency when the train is getTing closer.
Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:

hence, the frequencies for before and after tha train has past are
b) 472HZ, 408HZ
The object takes 0.5 seconds to complete one rotation, so its rotational speed is 1/0.5 rot/s = 2 rot/s.
Convert this to linear speed; for each rotation, the object travels a distance equal to the circumference of its path, or 2<em>π</em> (1.2 m) = 2.4<em>π</em> m ≈ 7.5 m, so that
2 rot/s = (2 rot/s) • (2.4<em>π</em> m/rot) = 4.8<em>π</em> m/s ≈ 15 m/s
thus giving it a centripetal acceleration of
<em>a</em> = (4.8<em>π</em> m/s)² / (1.2 m) ≈ 190 m/s².
Then the tension in the rope is
<em>T</em> = (50 kg) <em>a</em> ≈ 9500 N.