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Paraphin [41]
3 years ago
5

A long, fine wire is wound into a coil with inductance 5 mH. The coil is connected across the terminals of a battery, and the cu

rrent is measured a few seconds after the connection is made. The wire is unwound and wound again into a different coil with L = 10 mH. This second coil is connected across the same battery, and the current is measured in the same way. Compared with the current in the first coil, is the current in the second coil. ( Select all of them that applies)A- Twice as largeB- One-Fourth as largeC- UnchargedD- Half as largeE- Four times as large
Physics
2 answers:
Allisa [31]3 years ago
5 0

Answer:

unchanged

Explanation:

Let the voltage of the battery be V

Inductance L1 = 5 mH

Inductance L2 = 10 mH

consider resistance R of the circuit (wire, battery).

V = I R + L dI/dt

where, I is the current in the circuit and t is the time.

After few seconds of connection being made, the factor dI/dt is negligible. There is no change in the current flowing through the circuit. when inductor was just attached in the circuit, a current

Rom4ik [11]3 years ago
5 0

Answer:

Explanation:

The self inductance of the solenoid depends on the number f turns in the coil. As the battery remains same so the current remains same, but the number of turns changed so that the self inductance is changed.

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Answer:

induced current

Explanation:

intentionally manipulated.

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3 years ago
Describe the motion of a swing that requires 6 seconds to complete one cycle. What is its period and the frequency? Round to the
shutvik [7]

Period = 6 seconds and frequency = 0.167Hz .

<u>Explanation:</u>

We have , the motion of a swing that requires 6 seconds to complete one cycle. Period is the amount of time needed to complete one oscillation . And in question it's given that 6 seconds is needed to complete one cycle. Hence ,Period of the motion of a swing is 6 seconds . Frequency is the number of vibrations produced per second and is calculated with the formula of  \frac{1}{t} . SI unit of frequency is Hertz or Hz. We know that time period is 6 seconds so frequency =   \frac{1}{t}

⇒ frequency = \frac{1}{time}

⇒ frequency = \frac{1}{6}

⇒ frequency = 0.167Hz

Therefore , Period = 6 seconds and frequency = 0.167Hz .

7 0
3 years ago
A roller coaster car is going over the top of a 18-m-radius circular rise. at the top of the hill, the passengers "feel light,"
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The solution for this problem is:

If they feel 50% of their weight that means that the centripetal force is also 50% of their weight 1g - 0.5g = 0.5g 


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Then get the square root, the answer would be:
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8 0
3 years ago
Read 2 more answers
Two charges of equal magnitude Q are held a distance d apart. Consider only points on the line passing through both charges.For
Burka [1]

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A

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6 0
4 years ago
Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy
8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

Formula used for the radius of the n^{th} orbit will be,

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV

Energy of the second orbit in H atom .

E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV

Energy of the third orbit in H atom .

E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV

Energy of the fifth orbit in H atom .

E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV

Energy of the sixth orbit in H atom .

E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV

Energy of the seventh orbit in H atom .

E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV

During an absorption of energy electron jumps from lower state to higher state.So,  absorption will take place in :

1) n = 2 → n = 3

2) n=  5 → n = 6

Energy absorbed when: n = 2 → n = 3

E=E_3-E_2

E=(-1.51 eV) -(-3.40 eV)=1.89 eV

Energy absorbed when: n = 5 → n = 6

E'=E_6-E_5

E'=(-0.378 eV)-(-0.544 eV) =0.166 eV

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

4 0
3 years ago
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