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:
Acceleration, in m/s, of such a rock fragment = 
Explanation:
According to Newton's Third Equation of motion
Where:
is the final velocity
is the initial velocity
a is the acceleration
s is the distance
In our case:
So Equation will become:
Acceleration, in m/s, of such a rock fragment =
Answer:
If she stands on the North side of a river flowing to the East at 5 mph,
she must head towards the SouthWest to arrive on the South side of the river directly across from her starting point and we have
x^2 + 5^2 = 10^2 where x is her speed directly across the river
x = (75)^1/2 = 8.66 mph towards the South
sin theta = 5 / 10 = 1/2
She must angle the boat at 30 deg from straight South
Answer:
elements in the same column have the same number of neutrons. elements with similar mass are placed in the same column.
