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Softa [21]
4 years ago
15

Sources A and B emit long-range radio waves of wavelength 360 m, with the phase of the emission from A ahead of that from source

B by 90°. The distance rA from A to a detector is greater than the corresponding distance rB from B by 140 m. What is the magnitude of the phase difference at the detector?
Physics
2 answers:
Bumek [7]4 years ago
8 0

Answer:

The magnitude of the phase difference at the detector = 0.8692 rad

Explanation:

Given Data:

Wavelength (λ) = 360 m

Distance (x) = 140 m

The phase difference between the two phases can be calculated using the formula;

Ф = 2π/λ * x

   = (2*π/360) * 140

   =2.44 rad

Calculating the magnitude of the phase difference at the detector as;

ΔФ = 2.44 rad -90°(2π/360)

      = 2.44 - 1.5708

      = 0.8692 rad

 

chubhunter [2.5K]4 years ago
5 0

Answer:

5.5859rad

Explanation:

2pi/lambda(140)

=2pi/360x140

= 2.44rad

Thus total phase difference

Pi + 2.44rad

3.142+2.44

= 5.5859rad

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ANEK [815]

Answer:

a)

z=561.7

b)

z=214.1

Explanation:

L = inductance of the Inductor = 3.14 mH = 0.00314 H

C = capacitance of the capacitor = 5.08 x 10⁻⁶ F

a)

f = frequency = 55.7 Hz

Impedance is given as

z=\frac{1}{2\pi fC} - 2\pi fL

z=\frac{1}{2(3.14) (55.7)(5.08\times 10^{-6})} - 2(3.14) (55.7)(0.00314)

z=561.7

b)

f = frequency = 11000 Hz

Impedance is given as

z= - \frac{1}{2\pi fC} + 2\pi fL

z= - \frac{1}{2(3.14) (11000)(5.08\times 10^{-6})} + 2(3.14) (11000)(0.00314)

z=214.1

4 0
3 years ago
Which of the following types of light has the most energy to give to an electron?
IgorC [24]
I think it’s ultraviolet, if not i’m sorry!
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As an emergency vehicle approaches Bob and moves away from Jill, how does the actual frequency of the siren change? A) As an eme
Tamiku [17]
The correct answer is: 
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3 years ago
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A charge q1 of -5.00 x 10^-9 C and a charge q2 of -2.00x 10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium po
aleksandrvk [35]

Answer:

Explanation:

Let the equilibrium position of third charge be x distance from q₁.

Force on third charge due to q₁

= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²

Force on third charge due to q₂

= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

Both the force will act in opposite direction and for balancing , they should be equal.

9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

5  / x² = 2 / ( .4 - x )²

Taking square root on both sides

2.236 / x = 1.414 / .4 - x

2.236 ( .4 - x ) = 1.414 x

.8944 - 2.236 x = 1.414 x

.8944 = 3.65 x

x = .245 m

24.5 cm

So the third charge should be at a distance of 24.5 cm from q₁ .

4 0
3 years ago
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