The question is incomplete. The complete question is :
Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.
What is the frequency of the sound?
Solution :
Given :
The distance between the two loud speakers, 
The speaker are in phase and so the path difference is zero constructive interference occurs.
At the point 
, the speakers are out of phase and so the path difference is 
Therefore,
Thus the frequency is :
Hz
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.
Previous results tell us the speed (v) is given in terms of the coefficient of friction (k) and the radius of the curve (r) as
v = √(kgr)
v = √(0.20·9.8 m/s²·50 m)
= 7√2 m/s ≈ 9.90 m/s
In the experiment of free fall bob released a bag of mass 1 lb
so here we can say that initial speed of the bag is Zero
time taken by the bag to free fall is given as
t = 1.5 s
also the acceleration of free fall is given as
a = 9.8 m/s^2
now we will use kinematics equation here for finding the distance of free fall
so the bag will fall down by total distance of 11.025 m from its initial released position.
