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3 years ago

A positive charge of 1 µC is taken from points A & B such that VA-VB = 100 V. Then the

Physics

1 answer:

Lunna [17]3 years ago

4 0

Explanation:

It is given that,

Let Charge of $1\ \mu C$ is taken from points A & B such that $V_A-V_B=1000\ V$ .

We need to find the energy of charge. Electric potential is defined as the work done per unit of electric charge. So,

$W=(V_B-V_A)q\\\\W=-1000\times 10^{-6}\\\\W=E=-10^{-3}\ J$

So, the energy of charge decreases by $10^{-3}\ J$ . Hence, the correct option is (a).

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Who might be experiencing psychosis?

aleksandrvk [35]

Answer:

Explanation:

I think the answer is D, because of hallucinations

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3 years ago

_____ friction is the force that sliding objects experience

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Answer:

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Explanation:

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3 years ago

Item 19 Question 1 A Ferris wheel has a radius of 75 feet. You board a car at the bottom of the Ferris wheel, which is 10 feet a

sergey [27]

Answer:

$y = 104.4 ft$

Explanation:

As we know that we board in the car of ferris wheel at the bottom position

So we will have

final height of the car at angular displacement given as

$y = y_o + R + R sin(270 - 255)$

$y = y_o + R + R sin 15$

here we know that

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$R = 75 ft$

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$y = 10 + 75 + 75 sin15$

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3 years ago

Choose the law each sentence describes. This law relates a planet's orbital period and its average distance to the Sun. The orbi

hram777 [196]

These are the Kepler's laws of planetary motion.

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3 years ago

Read 2 more answers

Atomic physicists usually ignore the effect of gravity within an atom. To see why, we may calculate and compare the magnitude of

STatiana [176]

Answer:

$2.27\cdot 10^{49}$

Explanation:

The gravitational force between the proton and the electron is given by

$F_G=G\frac{m_p m_e}{r^2}$

where

G is the gravitational constant

$m_p$ is the proton mass

$m_e$ is the electron mass

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

$F_G=(6.67259\cdot 10^{-11} m^3 kg s^{-2})\frac{(1.67262\cdot 10^{-27}kg) (9.10939\cdot 10^{-31}kg)}{(3 m)^2}=1.13\cdot 10^{-68}N$

The electrical force between the proton and the electron is given by

$F_E=k\frac{q_p q_e}{r^2}$

where

k is the Coulomb constant

$q_p = q_e = q$ is the elementary charge (charge of the proton and of the electron)

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

$F_E=(8.98755\cdot 10^9 Nm^2 C^{-2})\frac{(1.602\cdot 10^{-19}C)^2}{(3 m)^2}=2.56\cdot 10^{-19}N$

So, the ratio of the electrical force to the gravitational force is

$\frac{F_E}{F_G}=\frac{2.56\cdot 10^{-19} N}{1.13\cdot 10^{-68}N}=2.27\cdot 10^{49}$

So, we see that the electrical force is much larger than the gravitational force.

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3 years ago