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Pavlova-9 [17]
3 years ago
11

An object is moving at 2.50 m/s [E]. At a time 3.00 seconds later the object is traveling at 1.50 m/s

Physics
1 answer:
babunello [35]3 years ago
7 0

Given parameters:

First velocity  = 2.50m/s

Time of travel = 3s

Second velocity  = 1.50m/s

Unknown:

The displacement during the first interval = ?

Velocity is the displacement of a body with time. Displacement is a distance move in a specific direction by a body.

    Velocity  = \frac{Displacement}{Time taken}

So;

      Displacement  = Velocity x Time taken

Now input the parameter for the first velocity and time of travel;

      Displacement  = 2.5 x 3  = 7.5m

The displacement id 7.5m

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A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
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a) E = 0

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The electric field for all points outside the spherical shell is given as follows;

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From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

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b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

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E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

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