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Two antennas located at points A and B are broadcasting radio waves of frequency 96.0 MHz, perfectly in phase with each other. T

earnstyle [38]

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

6 0

3 years ago

Read 2 more answers

Which object has the HIGHEST mass?

Arte-miy333 [17]

I think it would be A: gold ring

if you were to place all of these on a bed of water, the ring would be the first to sink.

5 0

3 years ago

The least count of stopwatch is 0.2s.The time of 20 oscillations of a pendulum was measured to be 25s.Find the percentage error

Svet_ta [14]

Answer:

0.8%

Explanation:

We are given;

Number of oscillations; n = 20

Time taken; t = 25 s

Formula for period of oscillation;

T = t/n = 25/20 = 1.25 s

We are told that the least count is 0.2 s. Thus, error is; ΔT = 0.2 s

percentage error in the measurement of time is given by;

(0.2/(20 × 1.25)) × 100% = 0.8%

4 0

3 years ago

A cyclist is riding a bicycle at a speed of 22 mph on a horizontal road. The distance between the axles is 42 in., and the mass

stealth61 [152]

Answer:

The shortest distance is $S = 24.86 ft$

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

The speed of the bicycle is $v_b = 22\ mph = 22 * \frac{5280}{3600} = 32.26 ft/s$

The distance between the axial is $d = 42 \ in$

The mass center of the cyclist and the bicycle is $m_c = 26 \ in$ behind the front axle

The mass center of the cyclist and the bicycle is $m_h = 40 \ in$ above the ground

For the bicycle not to be thrown over the

Momentum about the back wheel must be zero so

$\sum _B = 0$

=> $mg (26) = ma(40)$

=> $a = \frac{26}{40} g$

Here $g = 32.2 ft/s^2$

So $a = \frac{26}{40} (32.2)$

$a = 20.93 ft/s^2$

Apply the equation of motion to this motion we have

$v^2 = u^2 + 2as$

Where $u = 32.26 ft /s$

and $v = 0$ since the bicycle is coming to a stop

$v^2 = (32.26)^2 - 2(20.93) S$

=> $S = 24.86 ft$

5 0

3 years ago

The bending of wave crests as the reach shallow water is:

Blizzard [7]

Idk can u say it in a another way

5 0

3 years ago