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seropon [69]
3 years ago
13

In an LC circuit containing a 40-mH ideal inductor and a 1.2-mF capacitor, the maximum charge on the capacitor is 45mC during os

cillations. What is the maximum current through the inductor during the oscillations?
A) 3.7 A
B) 42 A
C) 2.5 A
D) 10 A
E) 6.5 A
Physics
1 answer:
GalinKa [24]3 years ago
3 0

Answer:

E) 6.5 A

Explanation:

Given that

L = 40 m H

C= 1.2 m F

Maximum charge on capacitor ,Q= 45 m C

The maximum current I given as

I = Q.ω

ω =angular frequency

\omega=\dfrac{1}{\sqrt{CL}}

By putting the values

\omega=\dfrac{1}{\sqrt{CL}}

\omega=\dfrac{1}{\sqrt{40\times 10^{-3}\times 1.2 \times 10^{-3}}}

ω  = 144.33 rad⁻¹

Maximum current

I = 45 x 10⁻³ x 144.33  A

I= 6.49 A

I = 6.5 A

E) 6.5 A

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sergeinik [125]
We use the formula v=ir where I is current, v is voltage, and r is resistance, we get that r=12/100 which is answer choice A
5 0
2 years ago
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A 240 V circuit has 20 A flowing through it. How much power is it using?
Delvig [45]

Explanation:

P = IV

V = 240

I = 20

P = ?

P = 240 × 20

P = 4800 watts

6 0
3 years ago
the distance between any two bodies is 10 M and the gravitational force between them is 3.2×10-⁹m. if the mass of one object is
cluponka [151]

Answer:

0.8 x 10^-9 kg

Explanation:

Given,

Distance ( R ) = 10 m

Force ( F ) = 3.2 x 10^-9 N

Mass ( m1 ) = 40 kg

To find : Mass ( m2 ) = ?

Formula : -

F = m1.m2 / R^2

m2 = FR^2 / m1

= 3.2 x 10^-9 x 10 / 40

= 3.2 x 10^-9 / 4

= ( 3.2 / 4 ) x 10^-9

m2 = 0.8 x 10^-9 kg

3 0
3 years ago
What is the value of n in the balmer series for which the wavelength is 410.2 nm.?
m_a_m_a [10]

The answer is n= 6.

What is Balmer series?

The Balmer series is the portion of the emission spectrum of hydrogen that represents electron transitions from energy levels n > 2 to n = 2. These are four lines in the visible spectrum. They are also known as the Balmer lines. The four visible Balmer lines of hydrogen appear at 410 nm, 434 nm, 486 nm and 656 nm.

For the Balmer series, the final energy level is always n=2. So, the wavelengths 653.6, 486.1, 434.0, and 410.2 nm correspond to n=3, n=4, n=5, and n=6 respectively. Since the last wavelength, 410.2 nm, corresponds to n=6, the next wavelength should logically correspond to n=7.

To solve for the wavelength, calculate the individual energies, E2 and E7, using E=-hR/(n^2). Then, calculate the energy difference between E2 (which is the final) and E7 (which is the initial). Finally, use lamba=hc/E to get the wavelength.

To learn more about emission spectrum click on the link below:

brainly.com/question/24213957

#SPJ4

3 0
2 years ago
A transverse wave on a string has an amplitude a. A tiny spot on the string is colored red. As one cycle of the wave passes by
aliya0001 [1]

Answer:

Option D) 4A

Explanation:

As the cycle of the wave passes by, the amplitude gives the longest journey when the spot travels from the undistributed position. During each cycle the spot travels "Four times" .

Considering one of this cycle, if it begins to travel from it's undistributed position , there would be four movements i.e

* Upward movement through distance A

*Downward movement through distance A

*Downward again through distance A

*Upward through distance A.

Then it would travel back to its undistributed position held

4 0
2 years ago
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