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matrenka [14]
3 years ago
5

According to Coulomb's law, when the distance between two point charges doubles, what happens to the electric force acting betwe

en the charges?
Physics
1 answer:
Dafna11 [192]3 years ago
3 0

Answer:

If the distance between charges is doubled, then the Force will become one-fourth.

F' = F/4

Explanation:

The Coulomb's Law gives the relation for the attractive or repulsive forces between two charges. The relation or formula is given as follows:

F = kq₁q₂/r²

where,

F = Attractive or Repulsive forces between the charges

k = Coulomb's Constant

q₁ = magnitude of 1st charge

q₂ = magnitude of 2nd charge

r = distance between charges

If we keep the magnitude of charges constant then the relation between Force an Distance between charges will be:

F α 1/r²

<u>So, it is clear from this relation that if the distance between charges is doubled, then the Force will become one-fourth.</u>

<u>F' = F/4</u>

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4 0
3 years ago
A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi
Dmitriy789 [7]

Answer:

a

 A =  0.081 \  m

b

The value is  u =  0.2569 \  m/s

Explanation:

From the question we are told that

   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

        The kinetic energy induced by the hammer  =  The energy stored in the spring

So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

3 0
2 years ago
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OLEGan [10]

Answer:

Pascal's law (also Pascal's principle[1][2][3] or the principle of transmission of fluid-pressure) is a principle in fluid mechanics given by Blaise Pascal that states that a pressure change at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere.[4] The law was established by French mathematician Blaise Pascal in 1653 and published in 1663.[5][6]

7 0
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Question 1 of 28
amm1812

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Explanation:

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Use the following half-life graph to answer the following question:
Temka [501]

Answer:

A 1.0 min

Explanation:

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From the graph in the problem, we see that the initial mass of the isotope at time t=0 is

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We see that this occurs at t = 1.0 min, so the half-life of the isotope is exactly 1.0 min.

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3 years ago
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