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BlackZzzverrR [31]
2 years ago
12

A metal sphere of radius 10 cm carries a charge of +2.0 μC uniformly distributed over its surface. What is the magnitude of the

electric field due to this sphere at a point 5.0 cm outside the sphere's surface? (k=1/4πϵ0=8.99×109 N · m2/C2) A metal sphere of radius cm carries a charge of μC uniformly distributed over its surface. What is the magnitude of the electric field due to this sphere at a point cm outside the sphere's surface? ( N · m2/C2) 4.0×109 N/C 4.0×107 N/C 8.0×107 N/C 4.2×106 N/C 8.0×109 N/C
Physics
1 answer:
frozen [14]2 years ago
3 0

Answer:

8.0\cdot 10^5 N/C

Explanation:

Outside the sphere's surface, the electric field has the same expression of that produced by a single point charge located at the centre of the sphere.

Therefore, the magnitude of the electric field ar r = 5.0 cm from the sphere is:

E=k\frac{q}{(R+r)^2}

where

k=8.99\cdot 10^9 N m^2C^{-2} is the Coulomb's constant

q=2.0 \mu C=2.0 \cdot 10^{-6}C is the charge on the sphere

R=10 cm = 0.10 m is the radius of the sphere

r=5.0 cm = 0.05 m is the distance from the surface of the sphere

Substituting, we find

E=(8.99\cdot 10^9 Nm^2 C^{-2})\frac{2.0\cdot 10^{-6} C}{(0.10 m+0.05 m)^2}=8.0\cdot 10^5 N/C

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Answer:

c

Explanation:

Your <em><u>wheels lose traction</u></em> on the road and your car <em><u>skids</u></em>

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2 years ago
If a young protostar with a disk is rotating and shrinking. how much faster is it rotating after its size has decreased by a fac
maks197457 [2]
In this system we have the conservation of angular momentum: L₁ = L₂
We can write L = m·r²·ω

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m₁ · r₁² · ω₁ = m₂ · r₂² · ω₂

The mass stays constant, therefore it cancels out, and we can solve for ω<span>₂:
</span>ω₂ =  (r₁/ r₂)² · ω<span>₁
     
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</span>ω₂ =  (4)² · ω<span>₁
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2 years ago
Substance X is placed in a container with substance Y. Both substances are fluids. Substance X initially sinks to the bottom of
Brut [27]

Answer: Option (A) is the correct answer.

Explanation:

Convection is a process in which heat transfers from a hotter substance to a colder substance.

As a result, the substance which is less dense will rise and the more denser substance will sink due to the influence of gravity.

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3 0
2 years ago
Read 2 more answers
A toroidal solenoid has an inner radius of 12.0 cm and an outer radius of 15.0 cm . It carries a current of 1.50 A . Part A How
tensa zangetsu [6.8K]

Answer:

The number of turns is  N  = 1750 \ turns

Explanation:

From the question we are told that

  The inner radius is r_i =  12.0 \  cm  =  0.12 \  m

   The outer radius is  r_o =  15.0 \  cm  =  0.15 \  m

   The current it carries is I =  1.50 \  A

    The magnetic field is  B  =   3.75 mT = 3.75 *10^{-3} \  T

   The distance from the center is d =  14.0 \ cm  =  0.14 \  m

Generally the number of turns is mathematically represented as

    N  =  \frac{2 *  \pi  * d  *  B}{ \mu_o *  r_o }

Generally  \mu_o is the permeability of free space with value  

    \mu_o  =  4\pi * 10^{-7} \ N/A^2

So

  N  =  \frac{2 *  3.142   * 0.14 *  3.75 *10^{-3} }{ 4\pi * 10^{-7}  * 0.15  }

  N  = 1750 \ turns

5 0
2 years ago
In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that
mars1129 [50]

Answer:

The maximum transverse speed of the bead is 0.4 m/s

Explanation:

As we know that the Amplitude of the travelling wave is

A = 3.65 mm

Now the speed of the travelling wave is

v_x = 13.5 m/s

now we know that distance of first antinode from one end is 27.5 cm

so length of the loop of the standing wave is given as

\frac{\lambda}{4} = 27.5 cm

\lambda = 110 cm

now we have

N = \frac{2L}{\lambda}

N = \frac{2(1.65)}{1.10}

N = 3

now we have

R = 2A sin(kx)

R = 2(3.65) sin(\frac{2\pi}{1.10}x)

R = 7.3 sin(1.82 \pi x)

now at x = 13.8 cm

R = 7.3 sin(1.82 \pi (0.138))

R = 5.18 mm

now we have

f = \frac{v}{\lambda}

f = \frac{13.5}{1.1}

f = 12.27 Hz

now maximum speed is given as

v_y = R\omega

v_y = (5.18 \times 10^{-3})(2\pi(12.27))

v_y = 0.4 m/s

4 0
1 year ago
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