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Ede4ka [16]
4 years ago
9

Two concentric conducting spherical shells produce a radially outward electric field of magnitude49,000 N/C at a point 4.10 m fr

om the center of the shells. The outer surface of the larger shell has aradius of 3.75 m. If the inner shell contains an excess charge of -5.30 μC, find the amount of chargeon the outer surface of the larger shell.
Physics
1 answer:
Korolek [52]4 years ago
6 0

To solve this problem it is necessary to apply the concepts related to the electric field according to the definition of Coulomb's law.

 The electric field is defined mathematically as a vector field that associates to each point in space the (electrostatic or Coulomb) force per unit of charge exerted on an infinitesimal positive test charge at rest at that point.

Mathematically this can be described as:

E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}

Where,

\epsilon_0 = permittivity of free space

r = Distance

q = Charge

E = Electric Field

Our values are given as,

E= 49.000N/C

r= 4.1m

\epsilon_0=8.854*10{-12}C^2N^{-1}m^{-2}

Replacing we have,

E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}

49000 = \frac{1}{4\pi (8.854*10{-12})}\frac{q}{(4.1)^2}

q= 9.16*10^{-5} C

q= 91.6\mu C

Therefore the amoun of charge on the outer surface of the larger shell is 91.6 \mu C

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vitfil [10]

Answer:

Final speed of the box after it has moved to x = 10 is given as

v = 3.35 m/s

Explanation:

As we know by work energy theorem that work done by all the forces is equal to the change in its kinetic energy

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so we have

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F x - \mu mg x = \frac{1]{2}mv^2

16.27 (10) - (0.15)(8)(9.8) (10) = \frac{1}{2}(8) v^2

162.7 - 117.6 = 4 v^2

v = 3.35 m/s

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4 0
3 years ago
Read 2 more answers
A basketball is tossed upwards with a speed of 5.0 m/s. ​We can ignore air resistance.
FinnZ [79.3K]

We have that  the maximum height reached by the basketball from its release point is

S=1.3m

From the question we are told

  • A basketball is tossed upwards with a speed of 5.0 m/s. ​We can ignore air resistance.  
  • What is the maximum height reached by the basketball from its release point?

Generally the Newtons equation for Motion  is mathematically given as

V^2=u^2+2as\\\\Therefore\\\\0=25-19.6*S\\\\S=\frac{25}{19.6}\\\\S=1.276

S=1.3m

Therefore

The maximum height reached by the basketball from its release point is

S=1.3m

For more information on this visit

brainly.com/question/23366835

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3 years ago
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