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Umnica [9.8K]
3 years ago
5

A charge q = 2.00 μC is placed at the origin in a region where there is already a uniform electric field \vec{E} = (100 N/C) \ha

t{i}E → = ( 100 N / C ) i ^. Calculate the flux of the net electric field through a Gaussian sphere of radius R = 10.0 \text{ cm}R = 10.0 cm centered at the origin. (ε0 = 8.85 × 10-12 C2/N ∙ m2)
Physics
1 answer:
tester [92]3 years ago
8 0

Answer:

Φ = 2.26 10⁶ N m² / C

Explanation:

The electric flow is

           Ф = E .ds = q_{int} / ε₀

Since we have two components for this flow. The uniform electric field and the flux created by the point charge.

The flux of the uniform electric field is zero since the flux entering is equal to the flux leaving the Gaussian surface

 

The flow created by the point load at the origin is

              Φ = q_{int} /ε₀

Let's calculate

              Φ = 2.00 10⁻⁶ /8.85 10¹²

              Φ = 2.26 10⁶ N m² / C

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Zolol [24]

Answer:

a) \omega = 0.421\,\frac{rev}{s}, b) \Delta \theta = 0.066\,rev, c) v = 0.966\,\frac{mm}{s}, d) a = 3.293\,\frac{mm}{s^{2}}

Explanation:

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\omega = \omega_{o} + \alpha\cdot \Delta t

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\omega = 0.421\,\frac{rev}{s}

b) The change in angular position is:

\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot  \alpha \cdot t^{2}

\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}

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a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )

a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}

a_{t} = 2.078\,\frac{mm}{s}

a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}

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a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}

a = 3.293\,\frac{mm}{s^{2}}

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Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

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