Do you have the answer choices ?
Answer:
a) Δφ = 1.51 rad
, b) x = 21.17 m
Explanation:
This is an interference problem, as they indicate that the distance AP is on the x-axis the antennas must be on the y-axis, the phase difference is
Δr /λ = Δfi / 2π
Δfi = Δr /λ 2π
Δr = r₂-r₁
let's look the distances
r₁ = 57.0 m
We use Pythagoras' theorem for the other distance
r₂ = √ (x² + y²)
r₂ = √(57² + 9.3²)
r₂ = 57.75 m
The difference is
Δr = 57.75 - 57.0
Δr = 0.75 m
Let's look for the wavelength
c = λ f
λ = c / f
λ = 3 10⁸ / 96.0 10⁶
λ = 3.12 m
Let's calculate
Δφ = 0.75 / 3.12 2π
Δφ = 1.51 rad
b) for destructive interference the path difference must be λ/2, the equation for destructive interference with φ = π remains
Δr = (2n + 1) λ / 2
For the first interference n = 0
Δr = λ / 2
Δr = r₂ - r₁
We substitute the values
√ (x² + y²) - x = 3.12 / 2
Let's solve for distance x
√ (x² + y²) = 1.56 + x
x² + y² = (1.56 + x)²
x² + y² = 1.56² + 2 1.56 x + x²
y2 = 20.4336 +3.12 x
x = (y² -20.4336) /3.12
x = (9.3² -20.4336) /3.12
x = 21.17 m
This is the distance for the first minimum
Answer:
400 N
Explanation:
Change of Kinetic Energy to Friction Wok
∆KE = W
½ x m x (v(5)² - v(3)²) = f x d
½ x 500 x (5² - 3²) = f x 10
250 x (25 - 9) = f x 10
25 x 16 = f
f = 400 N
Answer:
ΔL =0. 000312 m
Explanation:
Given that
At room temperature ( T = 25 ∘C) ,L= 1 m
So the length at 13.0 ∘C above room temperature
L=1.000312 m
So the change in length
ΔL = 1.000312 - 1.0000 m
ΔL =0. 000312 m
Answer:
Im just here for the points man sorry
Explanation:
