All categories

2 years ago

What capacitor in series with a 100ω resistor and a 18. 0 mh inductor will give a resonance frequency of 1000 hz?

1 answer:

5 0

Capacitor in series with a 100Ω resistor and a 18.0 mH inductor will give a resonance frequency of 1000 Hz is 1.38F

Resonance Frequency of a combination of Series LCR circuit is the frequency at which resonance is achieved in a circuit.

Resonance Frequency of a combination of Series LCR circuit is given by:

f = 1 / 2π $\sqrt{LC\\$

where, f is the resonance Resonance Frequency of a combination of Series LCR circuit

L is the Inductance of the inductor

C is the capacitance of the capacitor

Given,

f = 1000Hz

L = 18.0 mH = 0.018H

R = 100Ω

C = ?

On substituting the values in the above-mentioned formula:

1000 = 1 / 2π $\sqrt{0.018C}$

2π $\sqrt{0.018C}$ = 0.001

On solving, C = 1.38 F

Hence, the capacitance of the given capacitor is 1.38F.

Learn more about Resonant frequency here, brainly.com/question/13040523

#SPJ4

You might be interested in

A decorative plastic film on a copper sphere of 10 mm diameter in an oven at 750C. Upon removal from the oven, the sphere is sub

AlladinOne [14]

Answer:

3.26 secs

Explanation:

Diameter of sphere ( D )= 10 mm

T1 = 75°C

P = 1 atm

T∞ = 23°C

T2 = 35°c

Velocity = 10 m/s

<u>Determine how long it will take to cool the sphere to 35°C</u>

<em>Using the properties of copper and air given in the question</em>

Nu = 2 + (Re)^0.8 (Pr)^0.33

hd / k = 2 + ( vd/v )^0.8 (Pr)^0.33

∴ h ≈ 2594.7 W/m^2k

Given that :

(T2 - T∞) / ( T1 - T∞ ) = exp [ ( -hA / pv CP ) t ]

( 35 - 23 ) / ( 75 - 23 ) = exp [ - 2594.7 * 6 * t / 8933 * 387 * 10 * 10^-3 ]

= ln ( 12/52 ) = -1.466337069 = - 0.45032919 * t

∴ t ≈ 3.26 secs ( -1.466337069 / -0.45032919 )

6 0

3 years ago

12. What is the photoelectric effect, and what experimental evidence led scientists

anygoal [31]

Answer:

Photoelectric effect, phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation.

Explanation:

The effect is often defined as the ejection of electrons from a metal plate when light falls on it. In a broader definition, the radiant energy may be infrared, visible, or ultraviolet light, X-rays, or gamma rays; the material may be a solid, liquid, or gas; and the released particles may be ions (electrically charged atoms or molecules) as well as electrons. The phenomenon was fundamentally significant in the development of modern physics because of the puzzling questions it raised about the nature of light—particle versus wavelike behaviour—that were finally resolved by Albert Einstein in 1905. The effect remains important for research in areas from materials science to astrophysics, as well as forming the basis for a variety of useful devices.

8 0

3 years ago

How is 6.3 written in scientific notation? 6.3 mc022-1.jpg 100 63 mc022-2.jpg 10–1 6.3 mc022-3.jpg 101 63 mc022-4.jpg 100

hodyreva [135]

Not sure we can open those image files, but I think it would be 6.3*10^0

6 0

3 years ago

Read 2 more answers

Page worth 100 points please help

laila [671]

Answer:

a) the total circle of "initial" is Epot./Eel. and the total circle of "final" is Ekin.

b)intia: 0.5 * 50 * 4^2 = 400 J

final: 0 J = 0.5 * 50 * 0^2

Explanation:

c) in the initial phase the spring is compressed and so has Epot but then the Energie changes to be totally Ekin (in final phase) and so the Epot becomes 0

5 0

3 years ago

Read 2 more answers

Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of

EleoNora [17]

Answer:

tex]\lambda_{Be}[/tex] = 22.78 nm

Explanation:

Bohr's model for the hydrogen atom has been used by other atoms with a single electric charge by changing the number of charges by the charge of the new atom (atomic number)

$E_{n}$ = k e² / 2a₀ (1 /n²)

ao = h'² / k m e² h' = h/2πi

For another atom with a single electron in the last layer

a₀ ’= h’² / k m (Ze)²

a₀ ’= a₀ / Z²

Therefore, when replacing in the equation

$E_{n}$ = - Z² Eo/n²

E₀ = 13,606 eV

The transition occurs when the electron stops from one level to another

$E_{n}$ - $E_{m}$ = Z² E₀ (1 / n² - 1 / m²) = Z² ΔE

Let's relate this expression to the wavelength

c = λ f

E = h f

E = h c /λ

h c / λ = Z² ΔE

λ = 1 / Z² (hc / ΔE)

λ = 1 / Z² λ_hydrogen

Let's apply this last equation to our case

Lithium Z = 3

$E_{n}$ = - 9 Eo / n²

40.5 10-9 = 1/9 λ_hydrogen

Beryllium Z = 4

λ = 1/16 λ_hydrogen

Let's write our two equations is and solve

40.5 10-9 = 1/9 λ_hydrogen

tex]\lambda_{Be}[/tex] = 1/ 16 λ_hydrogen

40.5 10⁻⁹ = 1/9 (16 $\lambda_{Be}$ )

tex]\lambda_{Be}[/tex] = 40.5 9/16

tex]\lambda_{Be}[/tex] = 22.78 nm

6 0

3 years ago