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Sedaia [141]
3 years ago
5

Two charged spheres 10 cm apart attract each other with a force of 3.0 x 10^-6 N. What electrostatic force will result if both c

harges are doubled and the distance remains the same?
Physics
1 answer:
kherson [118]3 years ago
4 0

Answer:

The electrostatic force that will result if both charges are doubled and the distance remains the same is 6.0 * 10⁻⁶ N

Explanation:

Coulomb's law of  electricity states that the magnitude of the force of attraction or repulsion between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance of of separation between them.

In formula; F = KQ₁Q₂/d²

Using the formula for electrostatic force of attraction above to determine the force of between the two charged spheres

Let the charges be Q₁ and Q₂; distance of separation be d, K is a constant

Initially, F₁ = KQ₁Q₂/d² ---- (1)

F₁ = 3.0 * 10⁻⁶ N

when the charges are doubled, Q₁ = 2Q₁; Q₂ = 2Q₂; K and d remains constant

F₂ = 2KQ₁Q₂/d² ----(2)

Dividing equation (2) by (1) to find the ratio of their forces

F₂/F₁ = (2KQ₁Q₂/d²) / KQ₁Q₂/d²

F₂/F₁ = 2

Thus, F₂ is twice F₁.

Since F₁ = 3.0 * 10⁻⁶ N; F₂ = 2 * 3.0 * 10⁻⁶ N

F₂ = 6.0 * 10⁻⁶ N

Therefore, the electrostatic force that will result if both charges are doubled and the distance remains the same is 6.0 * 10⁻⁶ N

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An open pipe is 1.42 m long.
Varvara68 [4.7K]

Answer:

f(3) = 362.32 Hz

Explanation:

the formula of frequency of open pipe is

f(n) = (n+1)v/2L

n = 0,1,2,3,... (the order)

0 for the first

1 for the second

v = speed of sound

L = the lenght of pipe

f(3) = (2+1) * 343 / 2 * 1,42

f(3) = 3 * 343 / 2.84

f(3) = 362.324 Hz (or you can write 362.32 Hz

8 0
3 years ago
What could you do to change the volume of a gas
erma4kov [3.2K]

A gas always fills whatever container it's in.  You will get a lot of answers
to your question, with various technical discussions that involve pressure
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6 0
3 years ago
Two metal sphere each of radius 2.0 cm, have a center-to-center separation of 3.30 m. Sphere 1 has a chrage of +1.10 10^-8 C. Sp
Leokris [45]

Answer:

A)   V = -136.36 V , B)  V = 4.85 10³ V , C)  V = 1.62 10⁴ V

Explanation:

To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is

          V = k ∑ q_{i} / r_{i}

where q_{i} and r_{i} are the loads and the point distances.

A) We apply this equation to our case

          V = k (q₁ / r₁ + q₂ / r₂)

They ask us for the potential at the midpoint of separation

         r = 3.30 / 2 = 1.65 m

this distance is much greater than the radius of the spheres

let's calculate

         V = 9 10⁹ (1.1 10⁻⁸ / 1.65  + (-3.6 10⁻⁸) / 1.65)

         V = 9 10¹ / 1.65 (1.10 - 3.60)

         V = -136.36 V

B) The potential at the surface sphere A

r₂ is the distance of sphere B above the surface of sphere A

              r₂ = 3.30 -0.02 = 3.28 m

              r₁ = 0.02 m

we calculate

             V = 9 10⁹ (1.1 10⁻⁸ / 0.02  - 3.6 10⁻⁸ / 3.28)

             V = 9 10¹ (55 - 1,098)

             V = 4.85 10³ V

C) The potential on the surface of sphere B

      r₂ = 0.02 m

      r₁ = 3.3 -0.02 = 3.28 m

      V = 9 10⁹ (1.10 10⁻⁸ / 3.28  - 3.6 10⁻⁸ / 0.02)

       V = 9 10¹ (0.335 - 180)

       V = 1.62 10⁴ V

6 0
3 years ago
Which of the following characterizes a beta ray? Choose all that apply. is electromagnetic radiation is a product of natural rad
uranmaximum [27]

Explanation:

When a radioactive substance decays then the fast moving electrons emitted by it is known as beta ray. Basically, a number of beta particles are ejected by a beta ray.

Symbol of a beta particle is ^{0}_{-1}e. A beta ray is a natural decay of a radioactive element. As we know that opposite charges get attracted towards each other. So, a beta ray gets attracted towards a positively charged plate.

Therefore, we can conclude that following are the characterizes a beta ray:

  • a product of natural radioactive decay.
  • is attracted to the positively charged plate in an electric field.
  • is composed of electrons.
8 0
2 years ago
A(n) 83 kg fisherman jumps from a dock into a 139 kg rowboat at rest on the West side of the dock. If the velocity of the fisher
Anna71 [15]

here we can say that there is no external force on fisherman and dock

so here we will use momentum conservation theory

As per momentum conservation

initial momentum of fisherman + boat = final momentum of fisherman + boat

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

now we will have

83(3.1) + 139(0) = 83 v + 139 v

257.3 = 222v

v = 1.16 m/s

so the speed of boat and fisherman will be 1.16 m/s

3 0
3 years ago
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