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strojnjashka [21]
3 years ago
9

If the resistance in a wire is 273 Ω and the voltage is 79.6 volts, how much current is present

Physics
1 answer:
gtnhenbr [62]3 years ago
3 0

Answer:

0.292 A

Explanation:

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B, larceny because that's theft of personal property.

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What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from
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Quick Quiz 40.2 While standing outdoors one evening, you are exposed to the following four types of electromagnetic radiation: y
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The order of decreasing photon energy is FM radio, AM radio, yellow light, and microwaves.

Electromagnetic radiation: The electromagnetic field's waves, which are conveying electromagnetic radiant energy through space, are what make up electromagnetic radiation. It consists of X-rays, gamma rays, microwaves, infrared, light, and radio waves. These waves are all a component of the electromagnetic spectrum.

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2 years ago
Suatu gas memiliki volume 2 m³ dipanaskan dengan kondisi isobarik sehingga volumenya menjadi 5,5 m³ jika tekanan 4 atmosfer hitu
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3 years ago
A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri
GREYUIT [131]

Answer:

|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}

Now, substitute this into 'dF':

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)

Now we can integrate dF over the rod.

\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)

4 0
3 years ago
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