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pav-90 [236]
2 years ago
15

A very long wire carries a uniform linear charge density of 5 nC/m. What is the electric field strength 13 m from the center of

the wire at a point on the wire's perpendicular bisector? (ε 0 = 8.85 × 10-12 C2/N · m2)
Physics
1 answer:
kipiarov [429]2 years ago
5 0

Answer:

E=6.91 N/C

Explanation:

Given that

Linear Charge density ,λ = 5 nC/m

Distance ,R= 13 m

We know that formula for long wire to find electric field

E=\dfrac{\lambda }{2\pi \varepsilon _0R}

E=Electric field

R=Distance

εo=8.85 x 10⁻¹² C²/N.m²

λ=Linear Charge density

Now by putting the values

E=\dfrac{5\times 10^{-9}}{{2\times \pi \times 8.85\times 10^{-12}\times 13}}

E=6.91 N/C

Therefore the electric filed at distance 13 m will be 6.91 N/C

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This is the first stanza of the song:

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<span>Many a young maid lost her baubles to my trade </span>
<span>Many a soldier shed his lifeblood on my blade </span>
<span>The b*stards hung me in the spring of twenty-five </span>
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2 years ago
A quantity y is to be determined from the equation y=(px)/q^2
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Table 2.4 shows how the dispacement of a runner changed during a sprint race. Draw a dispacement-time graph to show this data, a
GalinKa [24]
4. Table 2.4 shows how the displacement of a runner changed
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8 0
1 year ago
A bag of potato chips contains 2.00 L of air when it is sealed at sea level at a pressure of 1.00 atm and a temperature of 20.0°
Genrish500 [490]

Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume V_{1} = 2.00\ L

Pressure P_{1}= 1.00\ atm

Pressure P_{2}= 70.0\ kPa

Temperature T_{1}= 20.0°C = 293\ K

Temperature T_{2}= 7.00°C = 280\ K

We need to calculate the volume at mountains

Using  gas law

\dfrac{PV}{T}=\ Constant

For both temperature,

\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}

Put the value into the formula

\dfrac{101.325\times2}{293}=\dfrac{70\times V_{2}}{280}

V_{2}=\dfrac{101.325\times2\times280}{293\times70}

V_{2}=2.766\ litre

Hence, The volume at mountains is 2.766 L.

5 0
2 years ago
A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k
rewona [7]

The kinetic energy of the small ball before the collision is

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Now is a good time to review the Law of Conservation of Energy:

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                     If it seems that some energy disappeared,
                     it actually had to go somewhere.
                     And if it seems like some energy magically appeared,
                     it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
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