Answer:
The lowest possible frequency of sound for which this is possible is 1307.69 Hz
Explanation:
From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.
First, we will determine his distance from the second speaker using the Pythagorean theorem
l₂ = √(2.00²+5.00²)
l₂ = √4+25
l₂ = √29
l₂ = 5.39 m
Hence, the path difference is
ΔL = l₂ - l₁
ΔL = 5.39 m - 5.00 m
ΔL = 0.39 m
From the formula for destructive interference
ΔL = (n+1/2)λ
where n is any integer and λ is the wavelength
n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.
Then,
0.39 = (1+ 1/2)λ
0.39 = (3/2)λ
0.39 = 1.5λ
∴ λ = 0.39/1.5
λ = 0.26 m
From
v = fλ
f = v/λ
f = 340 / 0.26
f = 1307.69 Hz
Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.
Answer:
The instantaneous velocity is the specific rate of change of position (or displacement) with respect to time at a single point (x,t) , while average velocity is the average rate of change of position (or displacement) with respect to time over an interval.Average velocity : Average velocity of a body is defined as the change in position or displacement (Δx) divided by time interval (Δt) in which that displacement occurs.
Instantaneous velocity : The instantaneous velocity of a body is the velocity of the body at any instant of time or at any point of its path .
velocity can be positive , negative or zero.
By studying speed and velocity we come to the result that at any time interval average speed of an object is equal or more than the average but instantaneous speed is equal to instantaneous velocity.
Answer:
v = 79.2 m/s
Solution:
As per the question:
Mass of the object, m = 250 g = 0.250 kg
Angle, 
Coefficient of kinetic friction, 
Mass attached to the string, m = 0.200 kg
Distance, d = 30 cm = 0.03 m
Now,
The tension in the string is given by:
(1)
Also
T = m(g + a)
Thus eqn (1) can be written as:





Now, the speed is given by the third eqn of motion with initial velocity being zero:

where
u = initial velocity = 0
Thus


Our Earth budget is how much energy we receive from the sun.
So the best answer is A