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:
-1.03 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity. The S. I unit of acceleration is m/s².
Mathematically, acceleration is expressed as
a = (v-u)/t ........................ Equation 1
Where a = acceleration, v = final velocity, u = initial velocity, t = time.
Given: u = 13.60 m/s, v = 7.20 m/s t = 6.2 s.
Substituting into equation 2
a = (7.20-13.60)/6.2
a = -6.4/6.2
a = -1.03 m/s²
Note: a is negative because, the hockey puck is decelerating.
Hence the average acceleration = -1.03 m/s²
The change in velocity from 30 m/s north to 40 m/s south is a change of 70 m/s south
Answer:
Explanation:
Density = Mass / Volume = 850 / 40*10*5 = 0.425 g /cm^3
Use pythagorean theorem
to find the opposite side, which is 7.3
so then you can just use inverse sinA=7.3/10 which equals 46.9 degrees