Answer:
Explanation:
magnetic filed, B = 0.65 T
initial diameter, d = 17.5 cm
final diameter, d' = 6.6 cm
time, t = 0.48 s
(a) According to Lenz's law, the direction of induced current is clockwise.
(b) Let e is the induced emf.
initial area, A = π r² = 3.14 x 0.0875 x 0.0875 = 0.024 m²
final area, A' = π r'² = 3.14 x 0.033 x 0.033 = 0.00342 m²
change in area, ΔA = A - A' = 0.024 - 0.00342 = 0.02058 m²
The magnitude of induced emf is given by
e = 0.65 x 0.02058 / 0.48
e = 0.028 V
(c) R = 2.5 ohm
i = e / R
i = 0.028 / 2.5
i = 0.011 A
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
That would be the atom I believe.
W = F * d
W = 10N * 2.5 m
W = 25 N m
So the answer you want is the third one down.
Mass of object/source
Gravity
Fg=mg
