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Oksana_A [137]
3 years ago
9

Two very small spheres are initially neutral and separated by a distance of 0.30 m. Suppose that 2.50 1013 electrons are removed

from one sphere and placed on the other. (a) What is the magnitude of the electrostatic force that acts on each sphere?
Physics
1 answer:
quester [9]3 years ago
4 0

Answer:

F = - 1,598 10⁻³ N

Explanation:

Electic strength is given by Coulomb's law

          F = k q₁ q₂  / r²

Where k is the Coulomb constant that is worth 8.99 10⁸ N m²/C², q₁ and q₂ are the charges and r is the distance that separates the electric charges

In this case the charge of the two spheres is the same and of a different sign since when you remove the charge of a sphere that was initially neutral, it is left with that charge removed but of the opposite sign

       q₁ = q₂ = 2.50 10¹³ electrons = 2.50 10¹³ 1.6 10⁻¹⁹

       q₀ = 4.0 10⁻⁶ C

Let's calculate

       F = - 8.99 10⁸ (4.0 10⁻⁶)² / 0.30²

       F = - 1,598 10⁻³ N

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A 2.20 kg mass is at the origin. A
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Answer:

0

Explanation:

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2 years ago
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A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
3 years ago
The Netherlands controlled ______ for centuries <br><br> A. Vietnam<br> B. Malaysia<br> C. Indonesia
gogolik [260]
The Netherlands controlled Indonesia for centuries!
4 0
3 years ago
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When AC Electric is producing 120 volts with 20 amps, What controls the the 120 volts at 60cycles a second?
lubasha [3.4K]

Answer:

Cycles per second is dependent on the construction of the alternator and the 120 volts is dependent upon the current and resistance in the circuit according to the ohms law.

Explanation:

We are given with AC of 120 volts, 20 amperes and 60 hertz frequency.

<u>According to the Ohm's law, we find its resistance:</u>

R=\frac{V}{I}

R=\frac{120}{20}

R=6\ \Omega

So, this 6 ohm resistance controls the current controls the magnitude of the AC current, while the frequency of the current remains constant and depends upon the construction and rotational speed of the armature of the alternator producing the current.

Here the value of frequency is the number of times the current changes its direction or the polarity in one second.

6 0
3 years ago
A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi
Serggg [28]

Answer:

The spring force constant is  k=243\ \frac{N}{m} .

Explanation:

We are told the mass of the ball is m=0.0555\ kg, the height above the spring where the ball is dropped is h=0.536\ m,  the length the ball compresses the spring is d=0.04897\ m and the acceleration of gravity is 9.8\ \frac{m}{s^{2}} .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy E_{mi} is equal to the final mechanical energy E_{mf} :

                                                    E_{mf}=E_{mi}

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

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If we manipulate the equation we have that:

                                                    k=\frac{2mgh}{d^{2} }

                                         k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}

                                              k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}

                                                     k=243\ \frac{N}{m}

                                                   

                             

5 0
3 years ago
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