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:
A fuse and circuit breaker both serve to protect an overloaded electrical circuit by interrupting the continuity, or the flow of electricity. ... Fuses tend to be quicker to interrupt the flow of power, but must be replaced after they melt, while circuit breakers can usually simply be reset.
Answer:
611.064 kJ
Explanation:
Given :
m = 200 mL = 200 g
Specific heat of ice = 2.06 J/g°C
Q = mcΔt
Δt = 0 - (-22) = 22
Q1 = 200 * 2.06 * 22 = 9064 J
Q2 = Melt 0 °C solid ice into 0 °C liquid water:
Q2 = m · ΔHf ; ΔHf = heat of fusion of water = 334j/g
Q2 = 200 * 334 = 66800 J
Q3 : Heat to convert from 0°C to 100°C
Q3 = mcΔt ; c = 4.19 J/g°C ; Δt = (100 - 0) = 100
Q3 = 200 * 4.19 * 100 = 83800 J
Q4: heat required to boil water to steam
Q = m · ΔHv
Hv = heat of vaporization of water = 2257 J/g
Q4 = 200 * 2257 = 451400 J
Total Q = Q1 + Q2 + Q3 + Q4
Q = 9064 + 66800 + 83800 + 451400
Q = 611,064 Joules
Q = 611.064 kJ
Answer:
Speed, u = 29.4 m/s
Explanation:
Given that, A ball thrown straight up climbs for 3.0 sec before falling, t = 3 s
Let u is speed with which the ball is thrown up. When the ball falls, v = 0
Using first equation of motion as :
v = u + at
Here, a = -g
So, u = g × t
u = 29.4 m/s
So, the speed with which the ball was thrown is 29.4 m/s. Hence, this is the required solution.
Is 7.35 because you subtic them and you're get the answer
