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ser-zykov [4K]
3 years ago
14

A positively charged particle initially at rest on the ground accelerated upward to 100m/s in 2.00s. If the particle has a charg

e-to-mass ratio of 0.100 C/kg and the electric field in this region is constant and uniform, what are the magnitude and direction of the electric field?
Physics
1 answer:
VashaNatasha [74]3 years ago
4 0

Answer:

Explanation:

From the question we are told that

     The initial velocity is  u  =  100 m/s

       The time taken is  t = 2.0 s

       The charge to mass ratio is  Q/m  =  0.100 C/kg

       

Generally the acceleration is mathematically evaluated as

              a  =   \frac{u}{t }

substituting values  

               a =   \frac{100}{2}

                a =  50 \ m/s^2

The electric field is mathematical represented as

             E = \frac{(a+g)}{Q/m}

substituting values

              E = \frac{(50+9.8)}{0.100}

              E = 598 \ N/C

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Answer:

E = k Q₁ / r²

Explanation:

For this exercise that asks us for the electric field between the sphere and the spherical shell, we can use Gauss's law

           Ф = ∫ E .dA = q_{int} / ε₀

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To solve these problems we must create a Gaussian surface that takes advantage of the symmetry of the problem, in this almost our surface is a sphere of radius r, that this is the sphere of and the shell, bone

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for this surface the electric field lines are radial and the radius of the sphere are also, therefore the two are parallel, which reduces the scalar product to the algebraic product.

         E A = q_{int} /ε₀

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          q_{int} = Q₁

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          E = k Q₁ / r²

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Answer:

Star X is much closer since it is at a distance 1 parsec from the Earth.

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The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth. This angle is gotten when the position of the object is measured in January and then in July according to the configuration of the Earth with respect to the Sun in those months.

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The parallax angle can be defined in the following way:

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p('') = \frac{1}{d}    (1)  

Equation (1) can be rewritten in terms of d:

d(pc) = \frac{1}{p('')} (2)

Equation (2) represents the distance in a unit known as parsec (pc).

Case of Star X (p('') = 1):

Using equation 2 the distance of star X can be known:

d(pc) = \frac{1}{1}

d(pc) = 1 pc

So, star X is at 1 parsec from Earth.

Case of Star Y (p('') = \frac{1}{2}):

d(pc) = \frac{1}{(\frac{1}{2})}

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So, star Y is at 2 parsecs from the Earth.

Hence, star X is much closer.

Reminder:

Notice that in equation 2 the distance is inversely proportional to the parallax angle, so if the parallax angle decreases, the distance increases.

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