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S_A_V [24]
3 years ago
8

A circuit element consists of a resistor with value 20Ω and inductor with value 10mH connected in series. A voltage of LaTeX: v(

t)=20\cos(1000t+45^{\circ})\text{V}v ( t ) = 20 cos ⁡ ( 1000 t + 45 ∘ ) V is applied across the circuit element. What is the apparent power of the circuit element? Answer in units of VA while omitting the unit. Hint: calculate the current first.
Physics
1 answer:
Flura [38]3 years ago
6 0

Answer:

8.97 Watt

Explanation:

Resistance, R = 20 ohm

Inductance, L = 10 mH

V(t) = 20 Cos (1000 t + 45°)

Compare with the standard equation

V(t) = Vo Cos(ωt + Ф)

Ф = 45°

ω = 1000 rad/s

Vo = 20 V

Inductive reactance, XL = ωL = 1000 x 0.01 = 10 ohm

impedance is Z.

Z = \sqrt{R^{2}+X_{L}^{2}}

Z = \sqrt{20^{2}+10^{2}}

Z = 22.36 ohm

V_{rms}=\frac{V_{0}}{\sqrt{2}}

V_{rms}=\frac{20}{\sqrt{2}} = 14.144 V

I_{rms}=\frac{V_{rms}}{Z}=\frac{14.144}{\sqrt{22.36}}=0.634 A

Apparent power is given by

P = Vrms x Irms

P = 14.144 x 0.634

P = 8.97 Watt

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Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to
jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x 10^{-8} J.

(b) The wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency of the photon is 2.47 x 10^{25} Hz.

Explanation:

Let;

E_{1} = -13.60 ev

E_{2} = -3.40 ev

(a) Energy of the emitted photon can be determined as;

E_{2} - E_{1} = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x 10^{-9})

E_{2} - E_{1} = 1.632 x 10^{-8} Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x 10^{-8} Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x 10^{-34} Js), c is the speed of light (3 x 10^{8} m/s) and λ is the wavelength.

10.20(1.6 x 10^{-9}) = (6.6 x 10^{-34} * 3 x 10^{8})/ λ

λ = \frac{1.98*10^{-25} }{1.632*10^{-8} }

  = 1.213 x 10^{-17}

Wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x 10^{-8}  = 6.6 x 10^{-34} x f

f = \frac{1.632*10^{-8} }{6.6*10^{-34} }

 = 2.47 x 10^{25} Hz

Frequency of the emitted photon is 2.47 x 10^{25} Hz.

6 0
2 years ago
If a photon has frequency = 2.00 x 1014s-1 and the speed of light = 3.00 x 108ms-1, then what is its wavelength?
butalik [34]

Answer:

The photon has a wavelength of 1.5x10^{-6}m

Explanation:

The speed of a wave can be defined as:

v = \nu \cdot \lambda (1)

Where v is the speed, \nu is the frequency and \lambda is the wavelength.

Equation 1 can be expressed in the following way for the case of an electromagnetic wave:

c = \nu \cdot \lambda (2)              

 

Where c is the speed of light.    

Therefore, \lamba\lambda can be isolated from equation 2 to get the wavelength of the photon.

\lambda = \frac{c}{\nu} (3)

\lambda = \frac{3.00x10^{8}m/s}{2.00x10^{14}s^{-1}}

\lambda = 1.5x10^{-6}m

Hence, the photon has a wavelength of 1.5x10^{-6}m        

<em>Summary:  </em>

Photons are the particles that constitutes light.

3 0
3 years ago
An umbrella tends to move upward on a windy day because _____.
masha68 [24]
E. all of the above

An umbrella tends to move upward on a windy day because _<span>A. buoyancy increases with increasing wind speed </span>
<span>B. air gets trapped under the umbrella and pushes it up </span>
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<span>D. a low-pressure area is created on top of the umbrella </span>

3 0
2 years ago
What is cocanve mirror?​
pochemuha

Answer:

A mirror that has a reflecting surface that is recessed inward is called concave mirror

8 0
2 years ago
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What is the relationship between electrostatic force and electric field?​
umka2103 [35]
Electrostatic attraction. The stronger the electric field, the greater the electrostatic force.
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