Answer:
from the midpoint we must move 1.34 m to any side and there is no sound
Explanation:
The sound waves are longitudinal waves that can have interference, for this to occur the difference in the path of the waves must be equal to an integer number of half wavelengths for the case from constructive interference and a semi-integer number for destructive interference.
Δd = 2n λ/2 constructive interference
Δd = (2n + ½) λ/2 destructive interference
Where n is an integer
At the midpoint between the two speakers the interference is always constructive, the two distances are equal, so the road difference is zero
Let's calculate the wavelength with the relationship
v = λ f
λ = v / f
λ = 344/641
λ = 0.537 m
We place the expression for the first destructive interference (no sound)
n = 1
Δd = (2 1 + ½) 0.537
Δd = 1.3425 m
This means that from the midpoint we must move 1.34 m to any side and there is no sound
I hope you get the answer soon dude
Answer:
. = 2574.08
Explanation:
given data
magnitude V1 = 80
magnitude V2 = 50
angle a = -37°
solution
= 80 k
= 50 cos{37} i - 50sin{37} k
so that here . is
. = 80 k . ( 50 cos{37} i - 50sin{37} k )
. = 80 k . ( 38.270 i + 32.176 k )
. = 2574.08
Answer:
No, false.
Explanation:
You may use it to power a light bulb if you desired. The reason it isn't is because it's inconsistent not predictable.
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m