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Anna [14]
3 years ago
9

A metal sphere of radius 2.0 cm carries an excess charge of 3.0 μC. What is the electric field 6.0 cm from the center of the sph

ere? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)
Physics
1 answer:
Nonamiya [84]3 years ago
5 0

Answer:

The electric field is  E = 7.5 *10^{6} \ N/C

Explanation:

From the question we are told that

    The radius of the metal sphere is  R = 2.0 \ cm  =  0.02 \ m

     The excess charge which the metal sphere carries is  q =  3.0 \mu C  =  3.0*10^{-6} \ C

      The distance of the position being to the center is D = 6.0 \ cm  = 0.06 \ m

       The coulomb constant is   k =9*10^{9} \  N \cdot m^2 /C^2

Generally the electric field is mathematically represented as  

        E = \frac{k *  q}{D^2}

substituting values

        E = \frac{9*10^{9} *  30.*10^{-6}}{(0.06)^2}

      E = 7.5 *10^{6} \ N/C

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\Delta x*\Delta p  \ \textgreater \ \hbar/2
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3 years ago
In his novel From the Earth to the Moon (1866), Jules Verne describes a spaceship that is blasted out of 12,000 yards/s. the Col
saw5 [17]

Answer:

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

Explanation:

Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration (a), measured in meters per square second, is estimated by this kinematic formula:

a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s } (1)

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\Delta s - Travelled distance, measured in meters.

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If we know that v_{o} = 0\,\frac{m}{s}, v = 10968\,\frac{m}{s} and \Delta s = 212.8\,m, then the acceleration experimented by the spaceship is:

a = \frac{\left(10968\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (212.8\,m)}

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The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

5 0
3 years ago
6.
S_A_V [24]

The distance in meters she would have moved before she begins to slow down is 11.25 m

<h3>LINEAR MOTION</h3>

A straight line movement is known as linear motion

Given that Ann is driving down a street at 15 m/s. Suddenly a child runs into the street. It takes Ann 0.75 seconds to react and apply the brakes.

To know how many meters will she have moved before she begins to slow down, we need to first list all the given parameters.

  • speed = 15 m/s
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speed = distance / time

Make distance the subject of the formula

distance = speed x time

distance = 15 x 0.75

distance = 11.25m

Therefore, the distance in meters she would have moved before she begins to slow down is 11.25 m

Learn more about Linear motion here: brainly.com/question/13665920

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algol [13]

Answer:

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