Explanation:
The vertical component = 10sin30° = 5.0N.
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:
6957.04N
Explanation:
Using
vf2=vi2+2ad
But vf = 0 .
So convert 50km/hr to m/s, and you need to convert 61 cmto m
(50km/hr)*(1hr/3600s)*(1000m/km) = 13.9m/s
61cm * (1m/100cm) = .61m
So n
0 = (13.9m/s)^2 + 2a(.61m)
a = 158.11m/s^2
So
using F = ma
F = 44kg(158.11m/s^2) = 6957.04N
Force = mass times acceleration
F = 21000 x 36.9 = 774900
Therefore, 774900N force is required.
