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prohojiy [21]
4 years ago
14

How does the electric force between two charged particles change if particle's charge is reduced by a factor of 3?

Physics
2 answers:
MAVERICK [17]4 years ago
3 0

The force between two charges is proportional to the product of the charges.

If only one of the charges is reduced by a factor of 3, then the force is reduced by a factor of 3.

If both charges are reduced by a factor of 3, then the force is reduced by a factor of 9.

mamaluj [8]4 years ago
3 0

Answer: It is reduced by a factor of 3.

Explanation:

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The newton's law of universal gravitation used to describe how a particle attracts every other particle in the universe.

The equation is given by,

F= G\frac{m_1 m_2}{r_2}

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m_1, m_2 = masses of the objects

r= is the distance between the center of their masses

G= Gravitational constant

For our problem we have defined that,

M_m = 7.24*10^{22}kg (mass of the moon)

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G= 6.671*10^{-11}Nm^2/kg^2

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We have then,

F_m = M\frac{6.67*10^-11*7.34*10^22}{3.84*10^8}

F_m = M*3.320*10^{-5} N

In the other hand we have the force on m-mass due to earth

F_E= MG = M*9.8N

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\frac{F_m}{F_E} = {M*3.320*10^{-5}}{9.8} = 3.387*10^{-6}

B) Suppose there is a group of young people surfing in the moonlight. They are directly under the moon. At this time the moon exerts its gravitational effect of the earth that causes the tide to rise. Around 6 hours later, when the earth has moved a quarter of the moon, the force on that point decreases, so the tide drops. However, after another 6 hours, people return and experience the same process. In this case the moon is not above them, but on the other side. This is because the moon having an orbit on the earth, generates an external force, similar to the previous one, but the earth reacts in the opposite way. It is like going in a car and turning it, all people will tend to get out of it, because a centrifugal force is experienced.

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3 years ago
A horizontal pipe 18.0 cm in diameter has a smooth reduction to a pipe 9.00 cm in diameter. If the pressure of the water in the
Rzqust [24]

Answer:

The rate of flow of water is 71.28 kg/s

Solution:

As per the question:

Diameter, d = 18.0 cm

Diameter, d' = 9.0 cm

Pressure in larger pipe, P = 9.40\times 10^{4}\ Pa

Pressure in the smaller pipe, P' = 2.80\times 10^{4}\ Pa

Now,

To calculate the rate of flow of water:

We know that:

Av = A'v'

where

A = Cross sectional area of larger pipe

A' = Cross sectional area of larger pipe

v = velocity of water in larger pipe

v' = velocity of water in larger pipe

Thus

\pi \frac{d^{2}}{4}v = \pi \frac{d'^{2}}{4}v'

18^{2}v = 9^v'

v' = 4v

Now,

By using Bernoulli's eqn:

P + \frac{1}{2}\rho v^{2} + \rho gh = P' + \frac{1}{2}\rho v'^{2} + \rho gh'

where

h = h'

\rho = 10^{3}\ kg/m^{3}

9.40\times 10^{4} + \frac{1}{2}\rho v^{2} = 2.80\times 10^{4} + \frac{1}{2}\rho (4v)^{2}

6.6\times 10^{4}  = \frac{1}{2}\rho 15v^{2}

v = \sqrt{\frac{2\times 6.6\times 10^{4}}{15\times 10^{3}}} = 8.8\ m/s

Now, the rate of flow is given by:

\frac{dm}{dt} = \frac{d}{dt}\rho Al = \rho Av

\frac{dm}{dt} = 10^{3}\times \pi (\frac{18}{2}\times 10^{- 2})^{2}\times 8.8 = 71.28\ kg/s

3 0
3 years ago
Read 2 more answers
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