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WINSTONCH [101]
3 years ago
5

Choose all correct sentences Group of answer choices The power is maximum when the value of the impedance is greater than the va

lue of the resistance. Resonance occurs when omega squared space equals 1 divided by space L C At resonance, the power is maximum and the impedance is minimum. At resonance, R squared space equals space (X subscript L minus X subscript c )squared The impedance Z is always larger than the resistance R.
Physics
1 answer:
Illusion [34]3 years ago
6 0

Answer:

True  b and c

Explanation:

In an RLC circuit the impedance is

         Z = \sqrt{[R^{2} + ( (wL)^{2} + (\frac{1}{wC})^{2} ]     }

examine the different phrases..

a) False. The maximum impedance is the value of the resistance  

b) True. Resonance occurs when  

              (wL)² + (1 / wC)² = 0

               w² = 1 / LC

c) True. In resonance the impedance is the resistive part and the power is maximum  

d) False. In resonance the inductive and capacitive part cancel each other out  

e) False. The impedance is always greater outside of resonance, but at the resonance point they are equal

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a = 9.8 m/s²

Explanation:

Acceleration due to gravity on Earth is constant, which is 9.8 m/s²

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A LED light source contains a 0.5-Watts GaAs (Eg =1.43 eV) LED. Assuming that 0.12% of the electric energy is converted to emiss
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Answer:

Explanation:

energy emitted by  source per second  = .5 J

Eg = 1.43 eV .

Energy converted into radiation = .5 x .12 = .06 J

energy of one photon = 1.43 eV

= 1.43 x 1.6 x 10⁻¹⁹ J

= 2.288 x 10⁻¹⁹ J .

no of photons generated = .06 / 2.288 x 10⁻¹⁹

= 2.6223 x 10¹⁷

wavelength of photon λ = 1275 / 1.43 nm

= 891.6 nm .

momentum of photon = h / λ  ;  h is plank's constant

= 6.6 x 10⁻³⁴ / 891.6 x 10⁻⁹

= .0074 x 10⁻²⁵ J.s

Total momentum of all the photons generated

= .0074 x 10⁻²⁵  x 2.6223 x 10¹⁷

= .0194 x 10⁻⁸ Js

b ) spectral width in terms of wavelength = 30 nm

frequency width = ?

n = c / λ  , n is frequency , c is velocity of light and λ is wavelength

differentiating both sides

dn = c x dλ / λ²

given dλ = 30 nm

λ = 891.6 nm

dn = 3 x 10⁸ x 30 x 10⁻⁹ / ( 891.6  x 10⁻⁹ )²

= 11.3 x 10¹² Hz .

c )

10 nW = 10  x 10⁻⁹ W

= 10⁻⁸ W .

energy of 50 dB

50 dB = 5 B

I / I₀ = 10⁵   ;   decibel scale is logarithmic , I is energy of sound having dB = 50 and  I₀ = 10⁻¹² W /s

I = I₀ x 10⁵

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= 10 x 10⁻⁸ W

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5 0
3 years ago
A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric
jeka94

Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

  • Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

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Answer:

d = 0.05 [m] = 50 [mm]

Explanation:

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E_{k}=F*d\\400 = 8000*d\\d = 0.05 [m] = 50 [mm]

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