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Westkost [7]
3 years ago
9

A circuit containing an inductor and a capacitor in series is designed to have a resonant frequency of 4511 Hz. If the inductor

has inductance 1.82 mH, then what is the required capacitance of the capacitor? Select one: a. 8.3e-4 Farads b.0.12 Farads c. 6.8e-10 Farads d. 2.7e-5 Farads e. 6.8e-7 Farads
Physics
1 answer:
OLEGan [10]3 years ago
3 0

Answer:

(e)6.835\times 10^{-7}F

Explanation:

At resonance we know that X_l=X_C

That is \omega L=\frac{1}{\omega C}

\omega ^2=\frac{1}{LC}

\omega =\frac{1}{\sqrt{LC}}

f=\frac{1}{2\pi \sqrt{LC}}

We have given resonance frequency f =4511 Hz and inductance L=1.82 mH

So 4511=\frac{1}{2\pi \sqrt{LC}}

LC=\frac{1}{4\pi ^2\times 4511^2}

LC=1.244\times 10^{-9}

C=\frac{1.244\times 10^{-9}}{1.82\times 10^{-3}}=0.6835\times 10^{-6}=6.835\times 10^{-7}F

So option e is the correct answer

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4 points
zubka84 [21]

(a) 25lx

(b) 11.11lx

<u>Explanation:</u>

Illuminance is inversely proportional to the square of the distance.

So,

I = k\frac{1}{r^2}

where, k is a constant

So,

(a)

If I = 100lx and r₂ = 2r Then,

I_2 = k\frac{1}{(2r)^2}

Dividing both the equation we get

\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(2r)^2}{k} \\\\\frac{I_1}{I_2} = 4\\\\I_2 = \frac{I_1}{4}\\\\I_2 = \frac{100}{4}  = 25lx

When the distance is doubled then the illumination reduces by one- fourth and becomes 25lx

(b)

If I = 100lx and r₂ = 3r Then,

I_2 = k\frac{1}{(3r)^2}

Dividing equation 1 and 3 we get

\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(3r)^2}{k} \\\\\frac{I_1}{I_2} = 9\\\\I_2 = \frac{I_1}{9}\\\\I_2 = \frac{100}{9}  = 11.11lx

When the distance is tripled then the illumination reduces by one- ninth and becomes 11.11lx

3 0
3 years ago
Goal posts at the ends of football fields are padded as a safety measure for players who might run into them. How does thick pad
dimaraw [331]
When looking at it from a physics standpoint, the thick padding decreases the amount of force that a player experiences. Force equals mass times acceleration where acceleration is distance over time. The thick padding increases the amount of time the player makes contact with the post which decreases force.
8 0
3 years ago
Read 2 more answers
In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P
Natasha_Volkova [10]

Answer

given,

distance of first satellite = 48,000 Km

distance of second satellite = 64,000 Km

orbital period = 6.39 day

Using equation of time period

  T = \dfrac{2\pi r^{3/2}}{\sqrt{Gm_{pluto}}}

now, from the above equation we can say that only variable is Time period and r is the radii of orbit.

from the first satellite

   \dfrac{T_{charon}}{r^{3/2}_{charon}}=\dfrac{T_{sat1}}{r^{3/2}_{sat1}}

   T_{sat1}=\dfrac{T_{charon}\ r^{3/2}}{r^{3/2}_{charon}}

   T_{sat1}=\dfrac{6.39\times (48000)^{3/2}}{19600^{3/2}}

   T_{sat1}=24.5\ days

for second satellite

   T_{sat2}=\dfrac{T_{charon}\ r^{3/2}_{sat2}}{r^{3/2}_{charon}}

   T_{sat1}=\dfrac{6.39\times (64000)^{3/2}}{19600^{3/2}}

   T_{sat1}=37.7\ days

7 0
3 years ago
Reading glasses of what power are needed for a person whose near point is 125 cm , so that he can read a computer screen at 54 c
Burka [1]

Answer:

1.11 dioptre

Explanation:

d_{i} = Distance of the image = - (125 - 2) = - 123 cm

d_{o} = Distance of the object = 54 - 2 = 52 cm

f = Focal length of the lens

Using the equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}

\frac{1}{52} + \frac{- 1}{123} = \frac{1}{f}

f = 90.1 cm

Power of the lens is given as

P = \frac{100}{f}

P = \frac{100}{90.1}

P = 1.11 Dioptre

3 0
3 years ago
An air-track cart is attached to a spring and completes one oscillation every 5.67 s in simple harmonic motion. at time t = 0.00
Masteriza [31]
The cart is moving by simple harmonic motion, and its position at time t is described by
x(t) = A \cos (\omega t)
where
A is the amplitude of the oscillation
\omega is the angular frequency

The amplitude of the oscillation corresponds to the maximum displacement of the spring, which corresponds to the initial position where the spring was released: 
A=0.250 m

The period of the motion is T=5.67 s, and the angular frequency is related to the period by
\omega =  \frac{2 \pi}{T}= \frac{2 \pi}{5.67 s} =1.11 rad/s

Therefore now we can calculate the position of the system at the time t=29.6 s:
x(29.6 s)=(0.250 m)\cos ((1.11 rad/s)(29.6 s))=+0.033 m
3 0
3 years ago
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