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:
14.8m
Explanation:
Given parameters:
Initial speed = 17m/s
Unknown:
Maximum height = ?
Solution:
At the maximum height, the final speed will be 0m/s;
We use of the kinematics equation to solve this problem.
V² = U² - 2gH
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity
H is the height
0² = 17² - (2 x 9.8 x h )
0 = 289 - (9.6h)
-289 = -19.6h
h = 14.8m
The force of gravity between two objects is given by:

where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation
In this problem, the mass of the object is

, while the Earth's mass is

. Their separation is

, therefore the gravitational force exerted on the object is
Answer:
The temperature is 90.4°C
Explanation:
See the attached for explanation
Given: Mass m = 60 Kg
Weight W = 96 N
Required: Acceleration due to gravity, g = ?
Formula: W = mg
g = W/g
g = 96 Kg.m/s²/60 Kg (note: this is the derive unit for Newton "N")
g = 1.6 m/s²