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igor_vitrenko [27]
3 years ago
11

A wire with resistance R is connected to the terminals of a 6.0 V battery. What is the potential difference between the ends of

the wire and the current I through it if the wire has the following resistances?
a) 1.0Ω
b) 2.0Ω
c) 3.0Ω
Physics
1 answer:
Bas_tet [7]3 years ago
7 0

Answer:

Potential difference = 6.0 V

I for 1.0Ω = 6 A

I for 2.0Ω = 3 A

I for 3.0Ω = 2 A

Explanation:

Potential difference (ΔV) = Current (I) x Resistance (R)

The potential difference is constant and equals 6.0 V, hence;

I = ΔV/R

When R = 1.0, I =6/1 = 6 amperes

When R = 2.0, I = 6/2 = 3 amperes

When R = 3.0, I = 6/3 = 2 amperes

<em>The potential difference is 6.0 V and the current is 6, 3, and 2 amperes for a resistance of 1.0, 2.0 and 3.0Ω respectively.</em>

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Answer:

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For n-=1 state hydrogen energy level is split into three componets in the presence of external magnetic field. The energies are,

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Here, E is the energy in the absence of electric field.

And

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Therefore,

\Delta E=2(9.3\times 10^{-24}J/T)\times 2.5 T\\\Delta E=46.5\times 10^{-24}J

Now,

Delta E=46.5\times 10^{-24}J(\frac{1eV}{1.6\times 10^{-9}J } )\\Delta E=29.05\times 10^{-5}eV\\Delta E\simeq29\times 10^{-5}eV

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irakobra [83]

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When the rocket fires the engines the gases leave at high speed and collide with the space station, transferring an impulse given by the expression

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When the spring, with the attached 275.0 g mass, is displaced from its new equilibrium position, it undergoes SHM. Calculate the
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Answer:

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Explanation:

Given that,

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\omega=\sqrt{\dfrac{k}{m}}

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T=1.33\ sec

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